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Chapter 6 — Optics

Class 12 · Physics

Overview

This unit on Optics explores the behaviour of light, how it travels, interacts with matter, and forms images. It covers geometrical optics — rays, reflection, refraction, mirrors, lenses, and optical instruments — and physical optics topics such as wave nature, interference, diffraction and polarization. The unit explains how laws of reflection and refraction arise, how lenses and mirrors form images, and how combinations of optical elements produce magnification. Students learn the principles behind human vision, microscopes, telescopes and cameras. In physical optics, the unit introduces superposition of waves, Young’s double-slit experiment, thin-film interference, diffraction patterns from single slits and gratings, and the concept of polarization and its applications. Understanding optics is essential for technologies such as spectacles, cameras, fibre-optic communications, lasers, and many modern instruments. Mastery of optics builds both conceptual reasoning and practical problem-solving skills, including ray diagrams, lens and mirror equations, interference fringe calculations, and resolving power. These topics provide foundations for further study in physics, engineering, and many applied sciences.

Learning Objectives

  • Describe the nature of light as rays and as waves and distinguish when each model applies.
  • Apply laws of reflection and refraction to solve ray-diagram and numerical problems for mirrors and lenses.
  • Use the mirror and lens formulae and magnification relations to locate images and compute sizes.
  • Explain formation of images by combinations of lenses and mirrors and calculate overall magnification.
  • Analyse single-slit and double-slit interference and diffraction patterns and calculate fringe positions.
  • Describe polarization of light and explain methods of producing and analysing polarized light.
  • Apply principles of thin-film interference to determine conditions for constructive and destructive interference.
  • Explain the working principles and resolving power of optical instruments such as microscopes and telescopes.

Topics in this chapter

17 topics · tap a topic title to jump straight to it.

💡1

Nature and propagation of light

Introduction to models: Light can be modelled in two complementary ways. The ray model treats light as straight-line rays; it works well when dimensions of objects and apertures are much larger than the wavelength. The wave model treats light as an electromagnetic wave: electric and magnetic fields oscillate perpendicular to the direction of propagation. This model explains interference, diffraction and polarization. In practical optics you will use the ray model for lenses and mirrors, and the wave model for phenomena where wavelengths matter.

Rectilinear propagation and rays: In a homogeneous transparent medium light travels along straight lines called rays. A ray marks the direction of energy flow and is perpendicular to the wavefront — a surface of constant phase. When rays encounter obstacles or boundaries, they can reflect or refract. Drawing ray diagrams helps locate images, predict shadows, and trace how optical devices work.

Wavefronts and Huygens’ principle: Huygens’ principle states that every point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront at a later time is the envelope of these wavelets. This construction is powerful: it explains how a plane wave remains plane, how spherical waves radiate from point sources, and provides a basis to derive the laws of reflection and refraction. It also shows how wave behaviour leads naturally to diffraction when obstacles have sizes comparable to wavelength.

Superposition and coherence: When two or more waves meet, the resulting displacement at any point is the algebraic sum of individual displacements — the principle of superposition. If the waves maintain a constant phase relationship (coherent), stable interference patterns form with bright and dark regions. Coherence has two aspects: temporal coherence (related to spectral bandwidth and coherence time) and spatial coherence (related to source size and angular extent). Lasers give long coherence lengths and high spatial coherence, making clear interference and holography possible.

Speed, frequency and wavelength: Light travels at speed c in vacuum; in a medium with refractive index n its speed is v = c/n. Frequency f remains unchanged when light crosses a boundary, while wavelength changes to λ = v / f = λ0 / n where λ0 is the vacuum wavelength. These relations are key to understanding refraction, dispersion (wavelength-dependent n), and thin-film interference.

Why both models matter: Selecting the correct model simplifies problem solving. Use rays to draw accurate image formation diagrams for mirrors, lenses and instruments; use waves to calculate fringe positions in interference and diffraction. Being fluent with both perspectives is essential for advanced optics and technological applications like imaging, communication and sensing.

📌 Examples
  • Drawing a ray path through two media using Snell’s law when a ray enters glass from air.
  • Explaining why shadows have sharp edges using rectilinear propagation.
  • Using Huygens’ principle to show why a plane wave remains plane after propagation over a short distance.
  • Calculating change of wavelength when light of 600 nm in air enters glass of n = 1.5.
🧮 Formulas
  1. v = c / n
  2. λ_medium = λ_vacuum / n
  3. Frequency f is unchanged across boundary: f1 = f2
📊 Visual ideas
Wavefronts and rays: draw parallel wavefronts with perpendicular rays; show secondary wavelets on Huygens’ construction.
Schematic of a plane wave crossing an interface with angles of incidence and refraction labelled.
💡2

Reflection of light: Plane and spherical mirrors

Laws of reflection: The basic rules are simple and powerful. The angle of incidence equals the angle of reflection measured from the normal to the surface, and the incident ray, reflected ray and normal lie in the same plane. These laws hold for smooth surfaces and are valid for plane and curved mirrors when applied locally to the tangent plane.

Plane mirrors: A plane mirror produces a virtual, upright image located the same perpendicular distance behind the mirror as the object is in front. Use two or three rays to construct the image: (1) a ray from the object to the mirror reflecting symmetrically about the normal, (2) a ray grazing parallel to the mirror which reflects back parallel, and (3) a ray to the mirror normal which retraces its path. The image size equals the object size and lateral inversion (left-right swap) occurs. Because the image is virtual it cannot be captured on a screen but is seen by an eye.

Spherical mirrors and principal points: A spherical mirror is part of a sphere with centre of curvature C and pole P. Concave mirrors (reflecting surface curved inward) can focus parallel rays to a real focus F, while convex mirrors (curved outward) always produce virtual images. For small-aperture mirrors (paraxial approximation) spherical mirrors behave predictably: rays close to the principal axis follow simple rules and form images described by mirror formulae.

Mirror formula and magnification: The mirror equation relates object distance u, image distance v and focal length f. Using a consistent sign convention 1/v + 1/u = 1/f, and magnification m = h' / h = -v / u. The focal length f relates to radius of curvature R by f = R/2 (paraxial approximation). These formulas let you compute image positions, sizes and types (real/virtual, erect/inverted) for various object placements relative to C and F.

Ray construction for spherical mirrors: Use principal rays: (i) a ray parallel to the principal axis reflects through focus, (ii) a ray through focus reflects parallel, (iii) a ray towards the centre C reflects back on itself. Combining two of these rays quickly locates the image. For concave mirrors, as the object moves from beyond C to between F and P, the image moves accordingly from between F and C to at infinity to virtual behind the mirror. Convex mirrors always give virtual, diminished and upright images behind the mirror.

Practical notes: Mirrors appear in many instruments, from shaving mirrors to telescopes. Understanding aberrations and paraxial limits is important for precise imaging, but first develop strong skills in using mirror equations and drawing accurate ray diagrams under the small-angle approximation.

📌 Examples
  • Locate image for an object placed at distance 3f from a concave mirror and compute magnification.
  • Explain image properties for object between F and P of a concave mirror.
  • Find focal length if the radius of curvature is 20 cm.
  • Ray diagram for object in front of a convex mirror and describe the image.
🧮 Formulas
  1. Mirror equation: 1/v + 1/u = 1/f
  2. Focal length and radius: f = R/2
  3. Linear magnification: m = h' / h = -v / u
📊 Visual ideas
Ray diagram for concave mirror showing object beyond C producing inverted real image between F and C.
Ray diagram for convex mirror showing virtual diminished image behind the mirror.
💡3

Refraction of light and Snell's law

Refraction: why light bends: Refraction is the change in direction of light as it crosses from one transparent medium to another. This happens because the speed of light changes between media. When a wavefront meets the interface at an angle, portions of the wavefront slow down first, causing the wavefront to change direction. The bending depends on the refractive indices of the two media.

Snell’s law and refractive index: The quantitative law is n1 sinθ1 = n2 sinθ2, where θ1 and θ2 are angles measured from the normal and n1, n2 are refractive indices. Refractive index n = c / v, so higher n means slower light and the medium is optically denser. Light entering a denser medium bends towards the normal; entering a rarer medium bends away.

Derivations and principles: Snell’s law can be given a wave explanation using Huygens’ principle or a variational explanation using Fermat’s principle of least time: light chooses the path that makes travel time stationary. These derivations reinforce the physical basis of refraction beyond memorising the formula.

Critical angle and total internal reflection: When light attempts to pass from denser medium n1 to rarer medium n2 (n1 > n2), beyond a certain angle θc there is no refracted ray; instead total internal reflection occurs. The critical angle satisfies sinθc = n2 / n1. This effect is not only interesting physics but also the operating principle of optical fibres and many types of prism-based devices where efficient light guiding is required.

Apparent depth and everyday examples: Refraction causes objects under water to appear closer to the surface than they are — the apparent depth can be calculated roughly as real depth divided by refractive index for normal viewing. Prisms exploit refraction to disperse white light into colours because n varies with wavelength (dispersion). Understanding how n depends on wavelength explains rainbows, chromatic aberration in lenses, and how to design antireflection coatings.

Problem solving tips: Always draw normals at interfaces, label angles and indices, and check whether angles are small or large to decide if approximations (like sinθ ≈ θ) may be used. Remember frequency remains unchanged across boundaries while wavelength and speed change. Practice tracing rays through multiple interfaces and calculating image shifts or angular deviations using Snell’s law and geometry.

📌 Examples
  • Calculate angle of refraction when ray in air (n=1) enters glass (n=1.5) at 30° incidence.
  • Find the critical angle for light going from glass (n=1.5) to air (n=1).
  • Explain why a coin in water appears raised (apparent depth calculation).
  • Using Snell’s law, trace a ray through a triangular prism and find deviation qualitatively.
🧮 Formulas
  1. Snell’s law: n1 sinθ1 = n2 sinθ2
  2. Critical angle: sinθc = n2 / n1 (for n1 > n2)
  3. Relative refractive index: m21 = n2 / n1
📊 Visual ideas
Ray diagram showing refraction at a plane surface with incident and refracted rays and normal.
Prism with incident ray, refracted ray, emergent ray and deviation angle.
🔍4

Refraction at spherical surfaces and lens-makers' formula

Refraction at a single spherical surface: When light refracts at a curved boundary between two media of refractive indices n1 and n2, the image formation follows a relation that generalises the plane-surface Snell’s relationship. For a spherical surface of radius R, with object distance u (measured from the pole) and image distance v, the formula is n1/u + n2/v = (n2 - n1)/R, provided sign conventions are used consistently. This equation helps locate images formed by one curved surface, for example a water drop acting as a lens.

Derivation idea and sign conventions: The formula comes from applying Snell’s law locally at the point of refraction and using simple geometry for small angles (paraxial rays). Positive and negative signs depend on the chosen convention; many problem sets use the Cartesian sign convention where distances measured against incident light are negative and radii of curvature are positive if centre is on outgoing side. The key is consistent use of signs throughout a calculation.

Thin lenses and two-surface refraction: A thin lens combines two spherical surfaces separated by a small thickness. For a thin lens in air the lens-maker’s formula relates the focal length f to the refractive index n and radii of curvature R1 and R2: 1/f = (n - 1)(1/R1 - 1/R2). This arises by applying the spherical-surface relation twice and neglecting the thickness term when the lens thickness is small compared to radii and object/image distances.

Understanding the lens-maker’s formula: The formula shows that focal length can be changed either by varying the lens curvature (R1, R2) or selecting different glass (n). A symmetric biconvex lens with equal radii has f = R / [2(n - 1)] approximately. A concave surface gives negative R in the sign convention, allowing concave or convex lenses to be described in the same formula.

Lens formula and magnification: For thin lenses, the imaging equation often used is 1/v - 1/u = 1/f (depending on sign choices) and magnification m = h'/h = -v/u. Positive focal length corresponds to converging lenses; negative to diverging. Students should be comfortable deriving image distances by combining lens-maker, thin-lens and magnification relations for a range of object positions and lens shapes.

Applications and practical tips: Use the lens-maker’s formula to design lenses of desired focal length, and use the spherical-surface formula for problems involving menisci, diaphragms or liquid surfaces. When dealing with thick lenses or complex systems, principal planes and effective focal lengths become important extensions of the thin-lens model.

📌 Examples
  • Compute focal length of a lens with n=1.5 and radii R1=20 cm, R2=-20 cm (bi-convex symmetric lens).
  • Use spherical surface formula to find image distance when air-to-glass refraction at a convex surface occurs.
  • Locate image for object placed at 30 cm from a converging lens of focal length 15 cm.
  • Compute magnification for the above lens and determine whether image is real or virtual.
🧮 Formulas
  1. Refraction at spherical surface: n1/u + n2/v = (n2 - n1)/R
  2. Lens-maker’s formula: 1/f = (n - 1)(1/R1 - 1/R2)
  3. Thin lens (image) formula: 1/v - 1/u = 1/f
  4. Magnification: m = h' / h = -v / u
📊 Visual ideas
Schematic of refraction at a single spherical surface showing object, image, centre of curvature and sign conventions.
Thin lens with principal axis, focal points and rays: parallel ray through focus and vice versa.
🔍5

Image formation and magnification by lenses

Principal rays and ray diagrams for thin lenses: Three convenient rays help locate images quickly. For a converging lens: (1) a ray parallel to the principal axis refracts through the far focal point, (2) a ray passing through the optical centre continues undeviated, (3) a ray passing through the near focal point emerges parallel to the axis. Their intersection gives the image. For diverging lenses use analogous rays and form virtual images by extending refracted rays backward.

Types of images based on object position: A converging (convex) lens shows different image behaviours as object distance changes relative to focal length f: if object is beyond 2f, image is real, inverted and smaller, located between f and 2f; at 2f object yields a real, inverted image at 2f of same size; between f and 2f the image is real, inverted and magnified beyond 2f; at f the rays emerge parallel and image is at infinity; inside f the lens forms a virtual, erect and magnified image on the same side as the object. Diverging (concave) lenses always form virtual, erect and diminished images.

Magnification formulas: Linear magnification m = h' / h = -v / u gives the signed ratio of image and object heights. Negative magnification indicates inversion when following sign conventions. Angular magnification is used for magnifiers and optical instruments: a simple magnifier’s angular magnification for relaxed eye M ≈ 25 cm / f, where 25 cm is a typical near point of distinct vision. For microscopes and telescopes magnification is the product of constituent lens magnifications.

Combination of thin lenses: Two thin lenses in contact behave as a single lens with focal length F given by 1/F = 1/f1 + 1/f2. When lenses are separated, treat the image from the first lens as the object for the second and apply successive imaging steps, taking care of sign conventions. For multi-element systems trace rays through each element and use effective focal length formulas where applicable.

Practical considerations and drawing skill: Accurate ray diagrams require scale drawings, marking principal points, focal lengths and axes. Practice diagrams to distinguish virtual versus real images and to find magnification. Numerical problems typically combine lens equations with geometry to compute positions and sizes, and instrument design problems use angular magnification relations to find eyepiece focal lengths or tube lengths for microscopes and telescopes.

📌 Examples
  • Find image position and magnification for object at 30 cm from a convex lens of f = 15 cm.
  • Determine image when object is at f/2 (inside focal length) for a convex lens.
  • Compute effective focal length for two lenses in contact with f1 = 10 cm and f2 = 20 cm.
  • Ray diagram for a magnifying glass viewing an object at near point with f = 5 cm.
🧮 Formulas
  1. Thin lens formula: 1/v - 1/u = 1/f
  2. Magnification: m = h' / h = -v / u
  3. Combined lenses in contact: 1/F = 1/f1 + 1/f2
  4. Angular magnification (simple microscope, relaxed eye): M = 25 cm / f
📊 Visual ideas
Ray diagrams for object positions: beyond 2f, at 2f, between f and 2f, and inside f showing image location and size.
Two thin lenses in contact represented by their principal planes and rays tracing through both.
🔭6

Optical instruments: Simple and compound microscopes

Simple magnifier (hand lens): A simple magnifier is a single converging lens used to view small objects. If the object is placed within the focal length, the lens forms a virtual, upright and magnified image. For the relaxed eye (final image at infinity), the angular magnification M ≈ 25 cm / f. For higher comfortable magnification the final image can be placed at the near point (25 cm) giving slightly different magnification formula. This device is handy and explains the principle of the eyepiece in compound instruments.

Compound microscope structure and operation: A compound microscope consists of an objective lens with a short focal length and an eyepiece (ocular). The objective, positioned close to the specimen, forms a real, magnified, and inverted intermediate image at a distance near the objective’s focal plane. The eyepiece then acts as a simple magnifier for this intermediate image, producing a much larger virtual image for the eye. The tube length L (distance between objective and eyepiece principal planes) and focal lengths determine magnification.

Magnification formula and design: The transverse magnification of the objective is approximately m_obj ≈ L / f_obj if the intermediate image lies near the focal plane of the eyepiece. The eyepiece’s angular magnification for a relaxed eye is M_eye ≈ 25 cm / f_eye. Total magnification is their product M_total = m_obj × M_eye. Thus, to increase overall magnification use a shorter focal length objective, a shorter eyepiece focal length, or increase tube length, while being mindful of practical limits like working distance and aberrations.

Resolution and numerical aperture: Magnification alone does not guarantee detail visibility; resolving power matters. The objective’s ability to resolve fine detail depends on numerical aperture NA and wavelength λ. According to Abbe, the minimum resolvable distance ~ λ / (2 NA). Increasing NA (using immersion oil with higher refractive index between specimen and objective) improves resolution by allowing larger acceptance angles. The objective lens design and quality of illumination (condenser) strongly affect image clarity.

Practical use and limits: Compound microscopes are essential for biology and materials science. In practice, mechanical stability, lens aberrations, and illumination control define real performance. Students should be able to compute magnifications using given focal lengths and tube lengths, draw ray diagrams showing intermediate and final images, and understand why very high magnification may not show additional detail if resolution is limited.

📌 Examples
  • Calculate total magnification for microscope with objective focal length 5 mm, eyepiece focal length 25 mm and tube length 160 mm.
  • Explain why an objective with shorter focal length increases magnification.
  • Describe how placing immersion oil increases resolving power.
  • Draw combined ray diagram showing intermediate image and final virtual image.
🧮 Formulas
  1. Objective magnification: m_obj ≈ L / f_obj
  2. Eyepiece magnification (relaxed eye): M_eye ≈ 25 cm / f_eye
  3. Total magnification: M_total = m_obj × M_eye
  4. Diffraction limit (qualitative): resolution ≈ λ / (2 NA)
📊 Visual ideas
Ray diagram for compound microscope showing objective producing real inverted image and eyepiece forming enlarged virtual image.
Schematic showing tube length L, objective and eyepiece focal lengths and image positions.
🔭7

Optical instruments: Astronomical and terrestrial telescopes

Purpose and basic principle: Telescopes gather light from distant objects and form magnified images so small angular separations become observable. They work by forming a real image of a distant object with the objective lens or primary mirror, then magnifying that image with an eyepiece. Because astronomical objects are effectively at infinity, telescopes are designed for angular magnification and high resolving power.

Keplerian (astronomical) refracting telescope: A Keplerian refracting telescope uses a converging objective lens to form a real inverted image at its focal plane. The eyepiece, also converging, is placed so that this image lies within the eyepiece focal length; the eyepiece produces a final virtual image at infinity for comfortable viewing. Angular magnification in normal adjustment (final image at infinity) is M = f_obj / f_eye. Keplerian telescopes provide an erect or inverted image depending on additional optics; astronomical use is fine with inverted images.

Galilean (terrestrial) telescope: The Galilean design uses a converging objective and a diverging eyepiece. The diverging eyepiece intercepts rays before they converge, producing an upright virtual image. Galilean telescopes give erect images and are short in length but have a smaller field of view and more limited performance compared to Keplerian designs.

Resolving power and aperture: The telescope’s ability to resolve fine detail depends on the diameter D of the objective. Diffraction sets a fundamental limit: the Rayleigh criterion gives the minimum resolvable angle θ_min ≈ 1.22 λ / D for a circular aperture. Larger apertures decrease θ_min and increase light-gathering power (proportional to D^2), letting the telescope observe fainter and closer objects.

Practical design considerations: Refractors suffer from chromatic aberration; achromatic doublets reduce this by combining glasses with different dispersions. Reflecting telescopes (using mirrors) avoid chromatic problems and scale more easily to large apertures. Mount stability, tracking, and atmospheric seeing are practical constraints. For numerical problems, students calculate magnification, focal lengths, image locations, and approximate resolving power using the formulas above.

📌 Examples
  • Compute angular magnification for telescope with objective f = 1200 mm and eyepiece f = 25 mm.
  • Compare image orientation for Keplerian and Galilean telescopes.
  • Estimate resolving power for an objective of diameter 200 mm for yellow light λ = 550 nm.
  • Explain why increasing magnification beyond a point makes images dimmer.
🧮 Formulas
  1. Telescope magnification: M = f_obj / f_eye (Keplerian, normal adjustment)
  2. Rayleigh criterion: θ_min ≈ 1.22 λ / D
  3. Light-gathering power ∝ D^2
📊 Visual ideas
Ray diagram for Keplerian telescope showing objective forming real image at focal plane and eyepiece forming virtual image at infinity.
Galilean telescope diagram showing eyepiece diverging lens producing erect image.
💡8

Interference of light: Principles and Young’s double-slit experiment

Interference basics and superposition: When two or more coherent waves meet at a point, the resultant electric field is the vector sum of individual fields. If phases add constructively the amplitude and intensity increase; if they add destructively they decrease. Stable interference requires coherence: fixed phase difference over time. Coherence can be achieved by splitting a single source into two paths so the waves remain phase-related.

Young’s double-slit experiment setup: In Young’s experiment a single monochromatic source illuminates two narrow, closely spaced slits. These slits act as coherent secondary sources producing overlapping waves on a distant screen. Under the approximation D >> d (screen distance much larger than slit separation), rays to a point on screen make approximately equal angles and path difference can be approximated by (d y)/D where y is lateral displacement from central maximum.

Conditions for fringes: Constructive interference (bright fringes) occurs when the path difference equals mλ (m integer). Destructive interference (dark fringes) occurs when path difference equals (m+1/2)λ. Using small-angle approximations, bright fringe positions y_m = m λ D / d and fringe spacing (fringe width) β = λ D / d. Central bright fringe corresponds to m = 0 and is located where path difference is zero.

Intensity and fringe contrast: For two equal-amplitude coherent waves the resultant intensity varies with phase difference δ as I = I_max cos^2(δ/2). Maximum intensity I_max occurs at δ = 2πm and minima at δ = (2m+1)π. Fringe visibility (contrast) depends on relative amplitudes and coherence; unequal amplitudes lower contrast while partial coherence blurs fringes. Temporal coherence relates to spectral width: broader spectra reduce fringe visibility for large path differences.

Practical uses and limitations: Young’s experiment is a fundamental demonstration of wave nature of light and provides a method to measure wavelengths. Real experiments must account for finite slit width (diffraction envelope), imperfect coherence, and alignment. Many optical instruments and techniques, including interferometers and thin-film optics, base their operation on the same interference principles.

📌 Examples
  • Calculate fringe separation if λ = 600 nm, slit separation d = 0.2 mm, screen distance D = 2 m.
  • Find position of 3rd bright fringe from central maximum using y = mλD/d.
  • Explain effect on fringe pattern if slit separation is doubled.
  • Compute intensity at a point with phase difference δ = π/2 for equal amplitudes.
🧮 Formulas
  1. Path difference ≈ d y / D (for D >> d and small angles)
  2. Condition for bright fringes: d sinθ = m λ
  3. Fringe width: β = λ D / d
  4. Intensity for two equal waves: I = I_max cos^2(δ/2)
📊 Visual ideas
Young’s double-slit setup: two slits separated by d, screen at distance D, central maximum and fringes labelled with spacing β.
Schematic showing path difference to a point on screen and angles.
🔬9

Interference in thin films

How thin-film interference arises: A thin film like oil on water or a soap bubble produces colours and patterns because light reflecting from its top and bottom surfaces interferes. The two reflected beams travel different optical paths and may experience phase inversions on reflection, so depending on thickness and wavelength the reflections reinforce or cancel.

Phase changes on reflection: A reflection from a boundary to a medium of higher refractive index produces a phase change of π (equivalent to half a wavelength). Reflection from a boundary to a lower-index medium does not add this π. When comparing two reflected beams, count the number of π phase shifts: if there is one inversion between them, the interference condition swaps compared to the case with no inversions.

Optical path difference and conditions: The extra path traveled inside the film is 2 n t cos r, where n is film refractive index, t thickness and r angle of refraction inside. For near-normal incidence cos r ≈ 1 so path difference ≈ 2 n t. If there is one phase inversion, constructive interference in reflected light occurs when 2 n t = (m + 1/2) λ (air wavelength), and destructive when 2 n t = m λ. If there are zero or two inversions, conditions swap: constructive for 2 n t = m λ. These rules let you compute which wavelengths reflect strongly and which cancel, producing the observed colours.

Angular dependence and colours: Because path difference depends on cos r, interference colours vary with viewing angle. Thicker regions of film favour longer wavelengths in constructive reflection; thin regions favour short wavelengths. The interplay of thickness variation and dispersion (n depends on λ) produces complex colourful patterns in soap bubbles and oil slicks.

Applications — antireflection coatings and sensors: Thin-film interference is used in antireflection coatings on lenses: a quarter-wavelength film of refractive index chosen appropriately causes reflections from the two surfaces to be out of phase and cancel for a target wavelength, reducing glare. Thin-film interference is also exploited in optical sensors and measurement techniques where small thickness changes cause measurable shifts in reflected colour or fringe position.

📌 Examples
  • Determine minimum thickness of an oil film (n = 1.4) to give constructive interference for λ = 600 nm (near-normal incidence) with one phase inversion.
  • Explain why soap bubbles show different colours at different places.
  • Design thickness for an antireflection coating for λ = 550 nm on glass (n_coating chosen suitably).
  • Calculate path difference for a 200 nm film at normal incidence with n = 1.33.
🧮 Formulas
  1. Optical path difference ≈ 2 n t cos r
  2. Constructive interference (with one phase inversion): 2 n t = (m + 1/2) λ
  3. Destructive interference (with one phase inversion): 2 n t = m λ
📊 Visual ideas
Thin film with two reflected rays shown: phase inversion at top surface indicated, path difference 2 n t labelled.
Diagram showing colours varying with thickness on a soap film and angles of observation.
🔋10

Diffraction: Single slit and resolving power

What is diffraction: Diffraction is the bending and spreading of waves when they pass an edge or aperture of size comparable to wavelength. It is inherently a wave phenomenon explained by Huygens’ principle: each point on an aperture becomes a secondary source and the resulting superposition produces characteristic intensity patterns in the far field. Diffraction becomes important when apertures are small or when precise resolution is required.

Single-slit diffraction pattern: For a slit of width a illuminated by monochromatic light, the far-field intensity shows a central maximum flanked by successive minima and weaker maxima. The condition for minima is a sinθ = m λ (m = ±1, ±2...). The central maximum is twice as wide as the adjacent maxima and carries most of the energy. The angular width of the central maximum is approximately 2λ/a, so narrower slits produce broader diffraction patterns.

Physical insight and energy distribution: Diffraction redistributes the light energy that would otherwise be concentrated in a geometrical spot. As aperture gets smaller, energy spreads over a wider angle, reducing peak intensity — an important consideration for imaging and illumination. The single-slit intensity I(θ) can be derived using integration across the aperture and has shape I(θ) = I0 (sin β / β)^2 where β = (π a sinθ)/λ.

Resolving power and circular apertures: Real optical systems have circular apertures (lenses, mirrors). The diffraction pattern for a circular aperture is the Airy pattern with a central disk and concentric rings. Rayleigh’s criterion sets the resolution limit: two point sources are just resolvable when the central maximum of one overlaps the first minimum of the other; the angular separation limit is θ_min ≈ 1.22 λ / D. Thus resolving power improves with larger aperture and shorter wavelength.

Practical implications: Diffraction limits the performance of microscopes, telescopes and cameras. Increasing magnification without increasing objective aperture does not reveal finer detail beyond the diffraction-limited resolution. In design, trade-offs exist: a larger aperture improves resolution but may introduce aberrations and weight. Students should be comfortable using the single-slit minima formula, Airy-disc criterion and understanding how aperture size influences intensity and sharpness.

📌 Examples
  • Find angle to first minimum for single slit width a = 0.1 mm and λ = 600 nm.
  • Compute resolving power for telescope of diameter 100 mm at λ = 550 nm using Rayleigh criterion.
  • Using grating equation, find angle for m=1 for grating with 5000 lines/cm for λ = 500 nm.
  • Explain why smaller aperture causes more spreading of light (wider diffraction pattern).
🧮 Formulas
  1. Single-slit minima: a sinθ = m λ (m = ±1, ±2, ...)
  2. Circular aperture (Rayleigh): θ_min ≈ 1.22 λ / D
  3. Grating equation: d sinθ = m λ
📊 Visual ideas
Intensity pattern for single-slit showing central maximum and minima positions.
Airy pattern schematic for circular aperture with central bright disc and rings.
11

Diffraction grating and spectral analysis

Diffraction grating principle: A diffraction grating has many equally spaced parallel slits or grooves. When monochromatic light is incident, light diffracted from different slits interferes constructively at specific angles satisfying d sinθ = m λ, where d is slit spacing and m is the order number. Because many slits contribute, principal maxima are very sharp and intense, making gratings excellent for precise wavelength measurements and spectral separation.

Orders and dispersion: Gratings produce multiple diffraction orders; the angular position of each order depends linearly on wavelength for small ranges, giving good dispersion. Angular dispersion dθ/dλ = m / (d cosθ) shows that dispersion increases with order m and with smaller slit spacing d. Linear dispersion on a detector at distance D is proportional to D times angular dispersion, so instrument design balances grating parameters and detector geometry.

Resolving power and spectral resolution: The resolving power R tells how well a grating separates two nearby wavelengths and is defined by R = λ / Δλ = m N where N is number of illuminated slits and m the order. High resolving power requires either large N (a large grating or wide illumination) or using higher orders, but higher orders may overlap and require order-selection filters. Gratings allow very fine discrimination of wavelengths and are widely used in spectroscopy.

Practical considerations and comparisons: Gratings provide near-linear dispersion and sharp lines; prisms give non-linear dispersion due to material dispersion. Gratings must be ruled or holographically produced with precise groove spacing; blaze angles can be engineered to increase efficiency in a preferred order. Overlap between orders may be mitigated by filters or using detectors selective to specific bands.

Applications and calculations: Use the grating equation to find angles for given wavelengths and orders; apply resolving power formula to determine minimum resolvable wavelength difference; compute dispersion to design spectrometers. Gratings are key components in laboratory spectrometers, astronomical spectrographs and devices that need accurate wavelength discrimination.

📌 Examples
  • Compute angle for first order (m=1) for λ = 500 nm with grating of 600 lines/mm.
  • Find resolving power for grating with 12000 illuminated lines in third order.
  • Calculate angular dispersion for m=2 at given θ and d.
  • Explain overlapping of spectral orders and how to avoid it using filters.
🧮 Formulas
  1. Grating equation: d sinθ = m λ
  2. Resolving power: R = λ / Δλ = m N
  3. Angular dispersion: dθ / dλ = m / (d cosθ)
📊 Visual ideas
Grating setup showing incident beam, diffracted beams at various orders with angles labelled.
Schematic of spectrum on a screen showing orders and possible overlap.
💡12

Polarization of light

What is polarisation: Polarisation describes the orientation of the electric field vector in an electromagnetic wave. Unpolarised light has electric field vectors in random directions perpendicular to the propagation direction. Polarised light has a definite orientation: linear polarisation has the E-field oscillating in one plane; circular or elliptical polarisation arises when two orthogonal components have a phase difference and appropriate amplitudes.

Production of polarised light: Common methods include polarising filters (which transmit one linear component), reflection at Brewster’s angle (where reflected light is plane polarised perpendicular to the plane of incidence), scattering (sky light is partially polarised), and birefringent crystals that split orthogonal components. Polaroid sheets are practical polarizers that absorb one component and transmit the orthogonal one.

Analyzing polarised light and Malus’ law: Malus’ law governs intensity transmitted through an analyser: I = I0 cos^2 θ where θ is angle between transmission axes. For unpolarised light passing through an ideal polariser intensity is halved. If a polarised beam passes through a second polariser at 90° (crossed), ideally no light is transmitted. Introducing a third polariser between crossed polarizers at an angle allows nonzero transmission due to rotation of the plane of polarization.

Circular and elliptical polarisation and retarders: When two perpendicular components are equal in amplitude and out of phase by ±π/2, the resultant is circular polarisation; unequal amplitudes or other phase shifts give elliptical polarisation. Wave plates (quarter-wave and half-wave plates) introduce controlled phase shifts between components and are used to convert linear to circular polarisation and to rotate polarisation planes, important in optical instrument design and communication systems.

Applications: Polarisation is used in glare reduction (sunglasses), liquid crystal displays (LCDs) rely on polarisation control, polarimetry measures material properties, and polarised light in microscopy enhances contrast for certain samples. Understanding polarisation and Malus’ law helps solve problems with intensity through multiple polarizers and designing devices that manipulate polarisation state.

📌 Examples
  • Use Malus’ law to compute transmitted intensity when analyser at 30° to polariser for I0 = 100 units.
  • Find Brewster’s angle for light going from air (n=1) to glass (n=1.5).
  • Explain effect of placing a quarter-wave plate between polariser and analyser to get circular polarization.
  • Determine transmitted intensity for unpolarised light through two ideal polarizers at 60°.
🧮 Formulas
  1. Malus’ law: I = I0 cos^2 θ
  2. Brewster’s angle: tan θ_B = n2 / n1
  3. Intensity through polariser for unpolarised light: I = (1/2) I_incident
📊 Visual ideas
Diagram of polariser and analyser with plane of polarisation and angle θ between them labelled.
Vector diagram showing two perpendicular components and phase difference producing circular polarization.
🔬13

Lasers and stimulated emission

Basic atomic processes: Atoms and molecules have discrete energy levels. When an electron drops from a higher level to a lower one it emits a photon. Spontaneous emission produces photons with random phases and directions. Stimulated emission occurs when an incoming photon of energy equal to the energy gap stimulates an excited atom to emit a second photon that has the same frequency, phase and direction as the stimulating photon. This mechanism provides coherent amplification of light in lasers.

Requirements for lasing: Three elements are necessary: a gain medium with suitable energy levels, population inversion where more particles occupy an excited state than the lower state, and an optical cavity that provides feedback. Pumping (optical, electrical or chemical) raises particles to the excited state. The cavity, formed by two mirrors, allows photons to pass back and forth, stimulating more emission and amplifying the beam. One mirror is partially transmitting to allow a controlled output beam.

Characteristics of laser light: Lasers are highly monochromatic (narrow spectral width), coherent (fixed phase relationship), directional (low divergence), and can have very high intensity. The coherence length relates inversely to spectral width: L_coh ≈ λ^2 / Δλ (qualitative). These properties make lasers ideal for interferometry, high-precision spectroscopy, holography, optical communication and cutting/welding applications.

Varieties of lasers: Lasers differ by gain medium: gas lasers (He–Ne), solid-state (ruby, Nd:YAG), semiconductor diode lasers, dye lasers and fiber lasers. Each has advantages in wavelength range, efficiency, power and application suitability. Mode structure (longitudinal and transverse) determines output frequency content and beam profile; advanced techniques like Q-switching and mode-locking produce high-energy pulses or ultrashort pulses respectively.

Applications and safety: Applications range from medical surgery and vision correction to barcode scanners, optical communications and research tools. Laser safety is important: power and wavelength determine hazard levels, especially to eyes. Appropriate eyewear, beam containment and procedural controls are essential when working with lasers.

📌 Examples
  • Explain why population inversion is necessary for laser action.
  • Describe how a He-Ne laser cavity produces a coherent beam.
  • List properties that distinguish laser light from ordinary light.
  • Explain the role of a partially reflecting mirror in a laser.
🧮 Formulas
  1. Photon energy: E = h ν = h c / λ
  2. Gain condition (qualitative): Amplification must overcome cavity losses for lasing action
  3. Relation between linewidth and coherence length: L_coh ≈ λ^2 / Δλ (qualitative)
📊 Visual ideas
Schematic of laser cavity with gain medium between two mirrors, showing emitted beam through partially transmitting mirror.
Energy level diagram showing stimulated and spontaneous emission and pumping leading to population inversion.
📈14

Holography and applications of interference

Holography records amplitude and phase: Holography goes beyond conventional photography by capturing both the amplitude and the phase of light scattered from an object. The essential idea is to create an interference pattern between the object wave (light scattered from the object) and a coherent reference wave. This pattern, recorded on a photographic plate or holographic medium, encodes the three-dimensional information of the object. When the developed hologram is later illuminated appropriately, it diffracts light to reconstruct the original wavefront, producing a three-dimensional image with depth cues such as parallax.

Recording process in detail: A coherent laser beam is split into two parts. One part illuminates the object; the light reflected or scattered from the object reaches the recording plate as the object beam. The other part, the reference beam, is directed to the plate without interacting with the object. On the plate the two wavefronts superpose and make an interference pattern of fringes. The fringe spacing and orientation depend on relative phase and path differences across the plate; these fine variations carry the phase information that allows reconstruction.

Stability and coherence requirements: High temporal coherence (narrow spectral width) ensures stable phase relations over the path differences involved; high spatial coherence ensures uniform phase across the beam. Lasers provide both qualities. During recording, mechanical stability is crucial—vibrations or air currents that change path lengths blur the fringes. Hence holography setups use vibration-isolated tables and short exposure times when possible.

Reconstruction and viewing: To view the hologram, illuminate it with the original reference beam (or a suitable substitute). The hologram diffracts the reference beam; the diffracted light reproduces the original object wavefront and creates a virtual image appearing to occupy the original object's space. Depending on recording geometry, both virtual and real images can be reconstructed. Transmission holograms are viewed with light passing through the plate; reflection holograms are viewed with light reflected back to the observer and may be visible in white light.

Applications and techniques: Holography has diverse uses: security features on banknotes and identity cards, high-density data storage ideas, holographic interferometry for measuring tiny deformations and vibrations, and holographic microscopy which allows numerical refocusing and phase contrast imaging. Holographic interferometry compares a reference hologram with a current hologram to reveal sub-wavelength displacements through fringe shifts; this is useful in engineering stress analysis and non-destructive testing.

Advantages and limitations: Holograms preserve depth information and parallax, giving a true 3D impression. However, they require coherent sources, a stable recording environment and careful processing. Modern digital holography uses sensors and computers to record and reconstruct wavefronts numerically, broadening practical applications by removing some chemical processing steps and allowing real-time reconstructions.

📌 Examples
  • Describe steps to record and view a simple laser hologram of a small object.
  • Explain why lasers are used in holography instead of ordinary lamps.
  • Give an application where holography provides benefit over photography.
  • Explain how holographic interferometry can measure small displacements.
🧮 Formulas
  1. Interference fringe condition: path difference = m λ (used in recording fringes)
  2. Phase difference δ = (2π / λ) × path difference
📊 Visual ideas
Setup diagram for hologram recording showing laser, beam splitter, object beam, reference beam and photographic plate.
Schematic of reconstructed wavefront from a hologram producing virtual and real images.
🔬15

Radiometry and photometry (basic concepts)

Why radiometry and photometry differ: Radiometry measures optical power and energy objectively across all wavelengths, using SI units such as watts for power. Photometry modifies radiometry to reflect human visual sensitivity: the eye does not respond equally to all wavelengths, so photometric quantities weight the spectral power by the eye’s response curve (the luminosity function) to produce measures in lumens and lux that match perceived brightness.

Key radiometric quantities: Radiant flux Φ (in watts) is the total optical power emitted, transmitted or received. Radiant intensity I_e (W/sr) measures power per unit solid angle. Irradiance E (W/m2) is power incident per unit area on a surface. Radiance describes directional brightness from a surface and combines area and angular distribution. These objective measures are used in physics and engineering when absolute power matters, such as laser output or detector calibration.

Photometric counterparts and conversion: Photometric quantities weight radiometric power by the photopic luminosity function V(λ). The luminous flux Φv (lumens) corresponds to radiometric power multiplied by spectral sensitivity and a constant: at 555 nm (where the eye is most sensitive) 1 W corresponds to 683 lm. For monochromatic light at 555 nm, Φv = 683 × Φe (with Φe in watts). For general spectra use Φv = ∫ 683 V(λ) Φe(λ) dλ integrating over wavelength. Illuminance (lux) equals luminous flux per unit area (lm/m2). Luminous intensity (candela) is luminous flux per unit solid angle (lm/sr).

Inverse-square law and practical illumination: For a point source emitting uniformly into space, irradiance falls off as E = Φ / (4 π r^2). This inverse-square relation is fundamental for lighting and sensor placement: doubling distance reduces irradiance by four. Real sources may be directional or extended; for such cases use solid-angle integrals or manufacturer's angular distribution data. In lighting design, lumens, lux and candela guide choices for fixtures, spacing and brightness levels.

Applications, measurement and safety: Radiometry is essential where absolute energy matters: laser power, solar panels, optical detectors and scientific instruments. Photometry is central to human-centred applications: room lighting, display brightness, and street lighting. Practical problems ask for irradiance at distances, conversion for monochromatic sources at 555 nm, and comparisons between luminous flux and intensity. When dealing with lasers or bright sources, remember safety considerations based on radiometric power and eye exposure limits.

📌 Examples
  • Compute irradiance at 2 m from a 10 W isotropic point source assuming all power is radiated uniformly.
  • Estimate luminous flux for 1 W monochromatic light at 555 nm using 683 lm/W.
  • Show how illuminance changes when moving twice as far from a small lamp.
  • Explain difference between luminous intensity (cd) and luminous flux (lm).
🧮 Formulas
  1. Irradiance (inverse-square): E = Φ / (4 π r^2) for isotropic point source
  2. Luminous flux at 555 nm: Φ_v (lm) = 683 × Φ_e (W) (for monochromatic 555 nm)
  3. General photometric conversion: Φ_v = ∫683 V(λ) Φ_e(λ) dλ
📊 Visual ideas
Graphical depiction of inverse-square law: point source and sphere of radius r with area 4πr^2 receiving power Φ.
Spectral sensitivity curve (qualitative) showing peak at 555 nm used in photometric weighting.
🔍16

Aberrations in lenses and their correction

What are optical aberrations: Aberrations are defects that make an optical system deviate from ideal image formation predicted by paraxial (small-angle) theory. They cause blurring, distortion, or colour fringes in images. Aberrations occur because real lenses and mirrors do not perfectly focus all rays to the same geometrical point. Recognising and correcting aberrations is central to high-quality optical design.

Common aberrations: Spherical aberration arises because spherical surfaces do not bring all rays to a common focus—marginal rays focus closer to the lens than paraxial rays, producing a blurred image. Chromatic aberration results from dispersion: different wavelengths have different refractive indices and thus different focal lengths. Coma produces comet-like tails for off-axis point sources; astigmatism transforms a point into lines depending on direction; field curvature means the best focus lies on a curved surface rather than a plane; distortion alters straight lines into pincushion or barrel shapes.

Correction methods: Several techniques reduce aberrations. Spherical aberration can be reduced by using aperture stops (blocking marginal rays), by designing aspheric surfaces that deviate from a sphere to correct ray paths, or by combining elements whose aberrations cancel. Chromatic aberration is corrected with achromatic doublets—pairing a crown and flint glass lens so two wavelengths (usually red and blue) focus at the same point. Apochromatic designs correct three wavelengths and further reduce residual colour errors.

Design trade-offs and practical considerations: Correcting aberrations often increases system complexity, size, weight and cost. Optical designers trade off performance across the field of view, wavelengths and desired image sharpness. For photographic lenses, stopping down aperture reduces many aberrations at the expense of diffraction effects; modern multi-element objectives and coatings reduce reflections and improve contrast.

Applications and examples: Telescope mirrors are often parabolic to avoid spherical aberration for parallel beams. Camera lens kits use multiple elements to correct aberrations over a wide field. Understanding aberrations helps predict image quality, choose correct lenses and interpret limitations when pushing magnification and resolution to extremes.

📌 Examples
  • Explain how an achromatic doublet reduces chromatic aberration for two wavelengths.
  • Describe effect of spherical aberration on image sharpness and how stopping down aperture improves sharpness.
  • Identify which aberration causes straight lines to bow near image edges (distortion).
  • Explain why aspheric lenses are more expensive but reduce spherical aberration.
📊 Visual ideas
Diagram showing focal shift of different wavelengths for chromatic aberration and how an achromat reduces this.
Sketch showing spherical aberration where marginal and paraxial rays meet at different points along axis.
🔭17

Applications of optics in technology and everyday life

Overview: Optics is the foundation of many common devices and modern technologies. Everyday items like eyeglasses, cameras and displays, and advanced systems such as fibre-optic networks, lasers in medicine and industry, and precision instruments in science, all rely on optical principles covered in this unit. Understanding how light is controlled, focused, dispersed or modulated helps explain their operation and limitations.

Vision and corrective lenses: The human eye forms a real image on the retina using a flexible lens; defects occur when the eye’s focal length does not match the eyeball length. Myopia (nearsightedness) is corrected with diverging lenses to move the focal point onto the retina; hypermetropia (farsightedness) with converging lenses. Presbyopia (loss of accommodation with age) is managed with reading glasses or multifocal lenses. These corrections use thin-lens formulas and magnification concepts to set lens powers.

Optical fibres and communications: Optical fibres guide light by total internal reflection within a core of refractive index higher than the cladding. Single-mode and multimode fibres trade off bandwidth and ease of coupling. Fibres enable high-bandwidth, long-distance data transmission with low loss, forming the backbone of modern internet infrastructure. Understanding critical angle, acceptance cone and dispersion are essential for designing and using fibre systems.

Cameras, imaging and sensors: Camera optics use lenses to form images on film or sensors. Aperture controls depth of field and exposure; focal length determines field of view and magnification. Diffraction limits and aberrations influence image sharpness. Digital sensors convert optical power to electrical signals; their sensitivity and pixel size interact with optical resolution and noise. Optical design blends physics and engineering to balance sharpness, brightness and portability.

Other applications: Interferometers measure small displacements, refractive index changes and surface irregularities. Polarisation is used in sunglasses to reduce glare, in LCD screens for display control, and in stress analysis through photoelasticity. Lasers are employed in surgery, cutting, surveying and barcode reading. Recognising the optical principles behind these devices allows practical problem solving and informed choices in everyday and professional contexts.

📌 Examples
  • Explain how eyeglasses correct myopia by using a diverging lens to move image onto retina.
  • Describe how a fibre-optic cable keeps light inside the core using critical angle concept.
  • Explain trade-off between aperture size and diffraction in camera design.
  • Give an example of using interferometry to measure small displacements.
🧮 Formulas
  1. Critical angle: sinθc = n2 / n1 (for n1 > n2)
  2. Relation between focus, focal length and object/image distances used in camera focusing: 1/v - 1/u = 1/f
📊 Visual ideas
Schematic of optical fibre cross-section showing core, cladding and guided rays reflecting at angles above critical angle.
Camera lens diagram showing aperture, lens, image sensor and principal distances.

Key Concepts

Ray optics
Model of light propagation treating light as straight-line rays useful when wavelength is much smaller than obstacles.
Wave optics
Model treating light as a wave to explain interference, diffraction and polarization.
Snell’s law
Relation n1 sinθ1 = n2 sinθ2 that governs refraction at a plane interface.
Total internal reflection
Phenomenon where light is completely reflected at a boundary when incidence exceeds critical angle.
Lens-maker’s formula
Relation 1/f = (n - 1)(1/R1 - 1/R2) connecting lens focal length with radii and refractive index.
Mirror equation
Relation 1/v + 1/u = 1/f linking object and image distances and focal length for mirrors.
Magnification
Ratio of image height to object height given by m = -v/u in thin-lens/mirror formulae.
Interference
Phenomenon where overlapping coherent waves produce regions of constructive and destructive superposition.
Young’s fringes
Interference fringes produced by two coherent slits with fringe spacing β = λD/d.
Diffraction
Spreading of waves when they pass through apertures comparable to wavelength, producing characteristic patterns.
Grating equation
Condition d sinθ = m λ for maxima produced by a diffraction grating.
Polarisation
Orientation property of transverse light waves describing direction of electric field oscillation.
Malus’ law
Formula I = I0 cos^2 θ giving transmitted intensity through an analyser at angle θ to polarisation direction.
Coherence
Property of waves that maintain constant phase difference necessary for stable interference.
Resolving power
Ability of an optical instrument to distinguish two close points; for circular aperture θ_min ≈ 1.22 λ/D.
Laser
Device producing coherent, monochromatic, and directional light by stimulated emission.
Holography
Technique recording both amplitude and phase of light waves to reconstruct three-dimensional images.
Aberration
Deviation from ideal image formation causing blurring, distortion or colour fringing.

Practice Questions

  1. An object 3 cm high is placed 30 cm from a concave mirror of focal length 15 cm. Find the position, nature and size of the image. / एक वस्तु जिसकी ऊंचाई 3 सेमी है वह 30 सेमी की दूरी पर एक अवतल दर्पण के सामने रखा गया है जिसका फोकल लंबाई 15 सेमी है। छवि की स्थिति, प्रकार और आकार ज्ञात कीजिए।
    Show answer

    Using mirror formula 1/v + 1/u = 1/f. Take u = -30 cm (object on incident side), f = -15 cm for concave mirror (using Cartesian sign convention) gives 1/v + 1/(-30) = 1/(-15). So 1/v = 1/(-15) + 1/30 = (-2 +1)/30 = -1/30, therefore v = -30 cm. Image is formed at 30 cm in front of mirror (real and inverted). Magnification m = -v/u = -(-30)/(-30) = -1 → image height h' = m h = -1 × 3 cm = -3 cm, so image is 3 cm high, inverted. / दर्पण सूत्र 1/v + 1/u = 1/f लागू करें। मानें u = -30 सेमी, फोकल लंबाई f = -15 सेमी (अवतल के लिए)। तो 1/v + 1/(-30) = 1/(-15) ⇒ 1/v = -1/15 + 1/30 = -1/30 ⇒ v = -30 सेमी। अतः छवि वस्तु के सामने 30 सेमी पर बनती है; यह वास्तविक और उल्टी है। आवर्धन m = -v/u = -(-30)/(-30) = -1 ⇒ छवि की ऊँचाई h' = -3 सेमी, यानी 3 सेमी और उल्टी।

  2. Light of wavelength 600 nm is incident on a double slit with slit separation 0.2 mm. The screen is 1.5 m away. Find the fringe separation. / तरंगदैर्घ्य 600 nm के प्रकाश को 0.2 mm छिद्र दूरी वाले द्वि-छिद्र पर डाला जाता है। परदा 1.5 m दूर है। फ्रिंज का विभाजन (fringe separation) ज्ञात कीजिए।
    Show answer

    Fringe width β = λ D / d. Here λ = 600 × 10^-9 m, D = 1.5 m, d = 0.2 × 10^-3 m. So β = (600×10^-9 × 1.5) / (0.2×10^-3) = (900×10^-9) / (0.2×10^-3) = (900/0.2)×10^-6 = 4500×10^-6 m = 4.5 mm. / फ्रिंज चौड़ाई β = λ D / d। λ = 600×10^-9 m, D = 1.5 m, d = 0.2×10^-3 m ⇒ β = (600×10^-9×1.5)/(0.2×10^-3) = 4.5×10^-3 m = 4.5 mm।

  3. A thin film of oil (n = 1.4) floats on water. What minimum thickness will produce constructive interference for normal incidence for wavelength 560 nm in air if the reflection from top surface undergoes phase inversion? / पानी पर तैरते तेल (n = 1.4) की एक पतली परत है। यदि ऊपर की सतह पर परावर्तन में चरण इनवर्जन होता है, तो वायु में तरंगदैर्घ्य 560 nm के लिए सामान्य आगमन पर न्यूनतम मोटाई कौन सी होगी ताकि रचनात्मक हस्तक्षेप हो?
    Show answer

    With one phase inversion, constructive condition: 2 n t = (m + 1/2) λ. Minimum non-zero thickness corresponds to m = 0: 2 n t = (1/2) λ ⇒ t = λ / (4 n). Substitute λ = 560 nm, n = 1.4: t = 560 nm / (4×1.4) = 560 / 5.6 nm = 100 nm. So minimum thickness = 100 nm. / एक चरण इनवर्जन की स्थिति में रचनात्मक: 2 n t = (m + 1/2) λ। न्यूनतम के लिए m = 0 ⇒ t = λ/(4 n)। अतः t = 560 nm / (4×1.4) = 100 nm।

  4. State Malus’ law and calculate transmitted intensity when polarised light of intensity 80 units passes through an analyser at 60°. / मालुस का नियम लिखिए और यदि ध्रुवीकृत प्रकाश जिसकी तीव्रता 80 इकाई है 60° पर एनालाइजर से गुजरता है तो पारित तीव्रता ज्ञात कीजिए।
    Show answer

    Malus’ law: I = I0 cos^2 θ, where I0 is incident intensity and θ is angle between transmission axes of polariser and analyser. Here I0 = 80, θ = 60°: cos 60° = 0.5, so I = 80 × (0.5)^2 = 80 × 0.25 = 20 units. / मालुस का नियम: I = I0 cos^2 θ। यहाँ I0 = 80, θ = 60°, cos60° = 0.5 ⇒ I = 80×0.25 = 20 इकाई।

  5. A diffraction grating has 5000 lines per cm. Find the angle for first order maximum for λ = 500 nm. / एक विवर्तन ग्रेटिंग में प्रति सेमी 5000 रेखाएँ हैं। तरंगदैर्घ्य 500 nm के लिए प्रथम क्रम का अधिकतम कोण ज्ञात कीजिए।
    Show answer

    Grating spacing d = 1 / (number per metre). 5000 lines/cm = 5000×100 = 5×10^5 lines/m so d = 1 / (5×10^5) = 2×10^-6 m. Grating equation d sinθ = m λ with m = 1: sinθ = λ / d = 500×10^-9 / 2×10^-6 = 0.25 ⇒ θ = arcsin(0.25) ≈ 14.48°. / d = 1/(5×10^5 m^-1) = 2×10^-6 m। sinθ = λ/d = 500×10^-9 / 2×10^-6 = 0.25 ⇒ θ ≈ 14.48°।

  6. Explain briefly why increasing the aperture diameter of a telescope improves its resolving power. / संक्षेप में समझाइए कि टेलिस्कोप के एपर्चर व्यास को बढ़ाने से उसका विभाजन-क्षमता (resolving power) क्यों सुधरती है।
    Show answer

    Resolving power limited by diffraction: for circular aperture Rayleigh criterion gives θ_min ≈ 1.22 λ / D. Increasing aperture diameter D decreases θ_min so two close point sources with smaller angular separation can be resolved. Larger aperture also increases light-gathering power improving visibility of faint details. Therefore increasing D improves resolving power. / विभेदन-शक्ति डिफ्रैक्शन से सीमित है। रैलेक्राइटेरियन के अनुसार θ_min ≈ 1.22 λ / D। D बढ़ाने से θ_min घटता है, अतः नज़दीकी दो स्रोतों को अलग देखा जा सकता है। बड़ा एपर्चर प्रकाश-संग्रहण भी बढ़ाता है जिससे मन्द वस्तुएँ स्पष्ट दिखती हैं।

  7. An object is placed 10 cm in front of a convex lens of focal length 15 cm. Describe the image formed. / एक समबाहु लेन्स जिसकी फोकल लंबाई 15 सेमी है के सामने 10 सेमी पर एक वस्तु रखी गई है। बन रही छवि का वर्णन कीजिए।
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    Object distance u = -10 cm (inside focal length since |u| < f). For a converging lens with f = +15 cm, thin lens formula 1/v - 1/u = 1/f gives 1/v - 1/(-10) = 1/15 ⇒ 1/v + 1/10 = 1/15 ⇒ 1/v = 1/15 - 1/10 = (2 - 3)/30 = -1/30 ⇒ v = -30 cm. Negative v indicates virtual image on same side as object, erect and magnified. Magnification m = -v/u = -(-30)/(-10) = -3 ⇒ image height is 3 times and inverted sign negative indicates erect? Note sign conventions: here v negative and u negative yield m = -v/u = -(-30)/(-10) = -3 → numerical magnification 3 and image is virtual and erect (convention consistency shows image virtual, upright and magnified by factor 3). / u = -10 cm, f = +15 cm. Using lens formula 1/v - 1/u = 1/f ⇒ 1/v + 1/10 = 1/15 ⇒ 1/v = -1/30 ⇒ v = -30 cm (virtual image on object side). The image is virtual, upright and magnified with linear magnification 3 (image three times larger).

  8. Why are lasers preferred in holography rather than ordinary light sources? / होलोग्राफी में सामान्य प्रकाश स्रोतों की अपेक्षा लेजर क्यों उपयोग किए जाते हैं?
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    Holography requires both high temporal and spatial coherence so that the object and reference beams maintain a stable phase relationship and produce clear interference fringes. Lasers provide high monochromaticity (long coherence length) and spatial coherence (collimated beam) necessary to record fine interference patterns. Ordinary lamps lack sufficient coherence and give washed-out or no stable fringes. Thus lasers are preferred. / होलोग्राफी के लिए वस्तु और संदर्भ तरंगों के बीच स्थिर चरण सम्बन्ध आवश्यक है; यह उच्च समयिक और स्थानिक सहसंबद्धता मांगता है। लेजर यह गुण प्रदान करते हैं — संकीर्ण तरंगदैर्घ्य और कोहेरेंट बीम — जबकि सामान्य ट्यूब वाले स्रोत में पर्याप्त कोहेरेंसी नहीं होती, इसलिए लेजरों को प्राथमिकता दी जाती है।

  9. A beam of unpolarised light of intensity 100 units passes through two ideal polarizers with their axes at 30° to each other. Find the transmitted intensity. / 100 इकाई तीव्रता वाला अप्रारोपित प्रकाश दो आदर्श पोलराइजर से गुजरता है जिनके अक्ष एक दूसरे से 30° पर हैं। पारित तीव्रता ज्ञात कीजिए।
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    Unpolarised light passing through first ideal polarizer emerges polarized with half the intensity: I1 = 100/2 = 50 units. Passing through analyser at angle θ = 30° gives I = I1 cos^2 θ = 50 × cos^2 30° = 50 × (√3/2)^2 = 50 × 3/4 = 37.5 units. / पहले पोलराइजर से अप्रारोपित प्रकाश का आधा ही निकलता है: I1 = 50। फिर Malus के अनुसार I = 50 cos^2 30° = 50×3/4 = 37.5 इकाई।

  10. Explain total internal reflection and give one application. / पूर्ण आंतरिक परावर्तन को समझाइए और एक अनुप्रयोग दीजिए।
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    Total internal reflection occurs when light tries to pass from a denser to a rarer medium at an angle of incidence greater than the critical angle θc, where sin θc = n2 / n1 (n1 > n2). For angles above θc no refracted ray exists and all light is reflected back into denser medium. An application is optical fibres: light is trapped in the core by repeated total internal reflections, enabling low-loss signal transmission over long distances. / पूर्ण आंतरिक परावर्तन तब होता है जब प्रकाश घने माध्यम से विरल माध्यम की ओर θ > θc पर आता है, जहाँ sin θc = n2/n1 (n1 > n2)। θc से अधिक परावर्तन ही होता है और प्रसरण नहीं। एक प्रसिद्ध अनुप्रयोग ऑप्टिकल फाइबर है जहाँ प्रकाश को कोर में पूर्ण आंतरिक परावर्तन द्वारा निर्देशित किया जाता है।

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