Overview
This unit on Light introduces the fundamental properties and behaviour of light that are essential in everyday life and technology. It covers the nature of light, how it travels, and how it interacts with surfaces and media. Students will learn the laws of reflection and refraction, how images are formed by plane and curved mirrors and by lenses, and how dispersion leads to colours. Practical phenomena such as total internal reflection, scattering and optical fibres are explained with applications. The unit also examines the human eye, eye defects and their correction, and basic optical instruments like the microscope and telescope. Understanding light helps explain vision, colour, instruments and communication systems, and provides a foundation for optics in higher studies and engineering. The topics develop observational skills, diagram drawing, ray tracing, and quantitative use of formulae such as the lens formula and refractive index, preparing students for board-level problems and real-world applications.
Learning Objectives
- Describe the nature of light and state its basic properties.
- Apply the laws of reflection to solve image-location problems for plane and curved mirrors.
- Apply Snell's law to solve problems on refraction and calculate refractive indices.
- Use lens formula and magnification formula to find image characteristics for thin lenses.
- Explain dispersion, scattering and how colours are formed or changed.
- Describe total internal reflection and explain its application in optical fibres and prisms.
- Explain the structure and functioning of the human eye and common eye defects with their corrections.
- Analyse simple optical instruments (microscope, telescope) and determine their magnifying power.
Topics in this chapter
17 topics · tap a topic title to jump straight to it.
Nature and Properties of Light
What is light?
Light is a form of energy that enables vision and carries information about the appearance of objects. In everyday situations, we study light using the ray model: light travels in straight lines called rays, and we draw these rays to predict shadows, images and paths through optical devices. For Class 10 purposes the ray picture gives clear, accurate predictions for reflection and refraction.
Sources and illumination
Objects that emit light are called luminous (e.g., the Sun, a lamp, a flame). Non-luminous objects become visible when they reflect light from luminous sources. The intensity or brightness at a point depends on how much light energy reaches a unit area there. For a point source the illumination decreases as distance increases: qualitatively we use this idea when explaining why a lamp brightens a nearby area more than a distant one.
Rectilinear propagation and shadows
Rectilinear propagation means that in a uniform medium light travels along straight lines. This is why sharp shadows form when the source is small or far away, and why pinhole cameras produce inverted images. When two light sources illuminate an object, parts of the screen may receive light from one source but not the other; this creates penumbra (partial shadow) and umbra (complete shadow).
Reversibility and superposition
Light rays obey reversibility: if a ray can travel from A to B along a path, it can travel from B to A along the same path. Superposition means different light beams add without changing each other’s paths (in geometric optics), allowing multiple images or patterns to overlap. These ideas are practical when tracing rays through lenses and mirrors.
Particle vs wave view (qualitative)
At this stage treat light primarily as rays, but be aware it also has wave-like properties (interference and diffraction) and particle-like behaviour (photoelectric effect studied later). Colour arises from wavelength — different wavelengths correspond to different colours in visible light. White light is a mix of wavelengths; prisms and droplets can separate these components to make a spectrum. Observations such as coloured patterns in thin oil films hint at wave effects, while straight-line behaviour and ray diagrams remain valid for many problems.
Practical observations students should try
Make a pinhole camera and note inversion; create shadows with single and double sources to see umbra/penumbra; observe brightness fall with distance. These simple experiments connect the ray model to real behaviour and prepare you to study reflection, refraction and image formation with confidence.
- Pinhole camera forming an inverted image of a distant object.
- Formation of umbra and penumbra using two torch lights and an object.
- Observation that a film of oil on water shows coloured patterns (thin-film hint).
Reflection of Light and Laws
What is reflection?
Reflection is the change in direction of light when it bounces off a surface back into the same medium. It is one of the most common interactions of light and is used in mirrors, periscopes and many optical instruments. Reflection can be regular (specular) from smooth surfaces, giving clear images, or diffuse from rough surfaces, scattering light in many directions and preventing clear images.
The two laws of reflection
The laws are simple and extremely useful. First, the incident ray, the reflected ray and the normal to the surface at the point of incidence all lie in the same plane. Second, the angle of incidence measured from the normal equals the angle of reflection. These rules let you draw accurate ray diagrams: mark the normal, measure the incident angle and then construct the reflected ray at the same angle on the other side.
Plane mirrors and image properties
Plane mirrors produce virtual images that appear to be as far behind the mirror as the object is in front. The image is upright and of the same size as the object but laterally inverted (left-right are swapped). Because image distance equals object distance, simple calculations and diagrams can locate the image precisely by drawing two or more reflected rays and extending them behind the mirror to their intersection.
Curved mirrors — qualitative
Curved mirrors (spherical sections) change the direction of reflected rays in ways that can form real images. A concave mirror can converge parallel rays to a focus, while a convex mirror causes them to diverge as if from a virtual focus behind the mirror. To analyze these mirrors, we use principal axis, pole, centre of curvature and principal focus — concepts introduced here and applied in later topics with mirror formula.
Ray construction and problem solving
For any mirror problem, draw the principal axis and identify the pole. For plane mirrors use two rays from a point on the object: one reflecting with equal angle and another via the pole (normal at the pole) to locate the image. For curved mirrors use principal rays: a ray parallel to the axis reflects through focus (or appears from focus for convex); a ray through centre reflects back on itself; a ray through focus emerges parallel. Intersections of reflected rays (or their backward extensions) give the image location and type (real or virtual), which determines magnification and orientation.
Practical applications
Mirrors are used in everyday life: plane mirrors in dressing, concave mirrors in torches and shaving mirrors to magnify, and convex mirrors in vehicle side mirrors for a wider field of view. Understanding reflection allows you to design and interpret these devices and to solve board-level ray tracing and numerical problems confidently.
- Ray diagram for a plane mirror showing object and virtual image at equal distances.
- Using the law of reflection to find direction of reflected ray when incidence angle is 30°.
- Using two rays to locate an image formed by a plane mirror of a small object.
- Angle of incidence (i) = Angle of reflection (r)
- For plane mirror: image distance = object distance
Spherical Mirrors: Concave and Convex
Geometry of spherical mirrors
Spherical mirrors are sections of a hollow sphere; concave mirrors have reflecting surfaces curved inward, convex mirrors have reflecting surfaces curved outward. Important points on the principal axis are the pole (P), the centre of curvature (C) which is the centre of the sphere, and the principal focus (F) which is midway between P and C for small apertures and paraxial rays. The distance PC is the radius of curvature (R) and PF is the focal length f where f = R/2 for a spherical mirror under paraxial approximations.
Principal rays and image formation in concave mirrors
To determine where an image forms, use three principal rays from a point on the object: (1) a ray parallel to the principal axis reflects through the focus; (2) a ray passing through the centre of curvature reflects back along itself (since it strikes the mirror normally); (3) a ray through the focus reflects parallel to the axis. Where the reflected rays meet (or their extensions meet) gives the image. Concave mirrors can produce a range of images: real and inverted when the object is beyond F, magnified when between F and C or between F and P, and virtual and erect when the object is between P and F.
Concave mirror characteristics by object position
Common positions and results are: object beyond C gives real, inverted and diminished image between F and C; object at C gives real, inverted image at C of same size; object between C and F gives real, inverted and enlarged image beyond C; object at F gives rays that emerge parallel and image at infinity; object between F and P gives virtual, erect and enlarged image behind the mirror. These cases are essential for sketching and solving numerical problems.
Convex mirrors and their features
Convex mirrors always produce virtual, erect and diminished images. Parallel rays appear to diverge from a virtual focus behind the mirror. Because the field of view is large, convex mirrors are commonly used as rear-view and security mirrors. Ray construction uses similar principal rays but reflected rays diverge and the virtual image is found by extending reflected rays backward behind the mirror.
Quantitative relations and sign conventions
The mirror formula 1/f = 1/v + 1/u and magnification m = v/u connect object distance u, image distance v and focal length f. Using the convention taught in class (typically object distances negative for real objects, real images positive etc.) allows calculation of image position and size. Careful use of sign conventions is important when solving problems involving concave and convex mirrors.
Practical uses
Concave mirrors are used in headlights, shaving mirrors, and telescopes to gather and focus light. Convex mirrors are used where a wide field of view is needed. Understanding ray diagrams and the mirror formula helps design and analyze these applications.
- Ray diagram for an object beyond C forming a real, inverted, smaller image between F and C.
- Ray diagram for object between F and P (concave) forming an enlarged virtual image.
- Ray diagram for convex mirror showing virtual, erect, and diminished image behind the mirror.
- Mirror formula: 1/f = 1/v + 1/u
- Magnification m = height of image / height of object = v/u
Refraction of Light and Refractive Index
What happens at a boundary?
When a light ray passes from one transparent medium into another, its speed changes because optical properties differ. This change in speed causes the light to bend at the interface; this bending is refraction. The direction of bending depends on whether the second medium is optically denser (higher refractive index) or rarer (lower refractive index) than the first. Light entering a denser medium bends toward the normal; entering a rarer medium it bends away from the normal.
Snell's law and refractive index
Snell's law gives the quantitative relation: n1 sin i = n2 sin r where n1 and n2 are refractive indices of the first and second media and i, r are angles of incidence and refraction measured from the normal. Refractive index n of a medium is defined as the ratio of the speed of light in vacuum (or air) to its speed in that medium: n = c/v. A larger n means light travels slower in that medium and refracts more strongly.
Relative and absolute refractive indices
The refractive index of medium A with respect to medium B is n_AB = v_B / v_A = sinθ_B / sinθ_A. If air or vacuum is the reference, we call it the absolute refractive index. For many problems air may be approximated as n ≈ 1.0. Refractive indices are dimensionless numbers and are measured experimentally by tracking incident and refracted angles and applying Snell's law.
Apparent depth and refraction at plane surfaces
Refraction causes objects under transparent media to appear shifted. For example, a coin at the bottom of a water beaker appears higher than its real position. This apparent raise is because light rays bend away from the normal as they leave water into air, so the brain interprets the rays as if they came in straight lines from a point nearer the surface. For small angles, apparent depth ≈ real depth / n (approximate relation) which helps estimate how much an object seems raised.
Refraction through glass slabs and lenses
A ray passing through a rectangular glass slab emerges parallel to its original direction but displaced sideways. For lenses and curved surfaces, refraction at two surfaces combines to form images; these systems are treated with lens formula and lens-maker relations. For refraction at a plane surface, use geometry together with Snell’s law to find image location given object position and refractive indices.
Everyday examples and measurement
Refraction explains the apparent bending of a straw in water, the sparkle of glass objects, and focusing in lenses of spectacles and cameras. Measuring refractive index can be done using prism deviation experiments or by measuring critical angle where the refracted ray grazes the boundary. Mastery of Snell’s law and refractive index calculations is vital for later lens and optical fibre topics.
- Using Snell's law to calculate the angle of refraction when light goes from air (n=1.0) to glass (n=1.5) at 30° incidence.
- Explaining why a straight stick looks bent when half in water.
- Calculating apparent depth if real depth is 2.0 m and n of water is 1.33.
- Snell's law: n1 sin i = n2 sin r
- Refractive index: n = c / v
- Apparent depth relation (qualitative)
Total Internal Reflection and Critical Angle
Understanding total internal reflection (TIR)
Total internal reflection is a special case of reflection that occurs when light attempts to pass from a denser medium into a rarer medium. Instead of being partly transmitted and partly reflected as usual, beyond a certain incident angle all the light is reflected back into the denser medium. This is not ordinary reflection from a surface but a complete confinement of light by the boundary due to wavefront behaviour and boundary conditions; for Class 10 it is sufficient to use geometric and Snell’s law reasoning to explain it.
Critical angle defined
The critical angle c is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°, so the refracted ray travels along the boundary. Using Snell’s law with r = 90°, we have n_denser sin c = n_rarer sin 90° = n_rarer, hence sin c = n_rarer / n_denser. Critical angle exists only when n_denser > n_rarer; there is no critical angle when light tries to go from rarer to denser medium.
Behaviour as angle increases
For incidence angles less than c, part of the ray is refracted and part reflected. At exactly c the refracted ray grazes the surface. For incidence angles greater than c there is no refracted ray; all light is reflected internally. This behaviour is predictable using Snell’s law and the sine function exceeding unity for impossible refraction angles.
Applications of TIR
TIR underpins optical fibres, prisms used in binoculars and periscopes, and some medical instruments. Optical fibres use a core with a slightly higher refractive index surrounded by cladding to ensure internal rays meet the boundary at angles greater than the critical angle so they remain guided. Prisms such as right-angled prisms use TIR to turn light beams without loss from mirrored surfaces and to preserve image brightness and quality.
Practical demonstrations and safety
Using a semi-circular glass block or acrylic block, students can vary incidence angle and observe the refracted ray shortening until it disappears and only reflected ray remains — demonstrating TIR and enabling measurement of critical angle. In fibre demonstrations, gentle bends still guide light because internal reflections continue, but sharp bends may cause some rays to strike below the needed angle and leak out. Avoid shining high-power lasers into fibres or eyes during demonstrations.
Quantitative example
For glass (n≈1.5) to air (n≈1.0), sin c = 1/1.5 ≈ 0.6667, so c ≈ 41.8°. Rays in glass striking at angles above 41.8° relative to the normal will be totally internally reflected, which is the design principle for many optical devices.
- Calculate critical angle when light goes from glass (n=1.5) to air (n=1.0): sin c = 1/1.5 so c ≈ 41.8°.
- Explain how light remains inside an optical fibre using total internal reflection.
- Ray diagram showing a ray inside a glass slab undergoing TIR at the top surface.
- Critical angle: sin c = n1 / n2 (for n2 > n1)
Refraction through Lenses and Lens Formula
What is a lens?
A lens is a transparent optical element bounded by two curved surfaces that refract light to form images. Lenses can be converging (convex) or diverging (concave). For thin lenses, thickness is small compared to radii of curvature and focal length, so we can use simple formulas and ray rules. The principal axis passes through the centre of the lens, and the optical centre is a point through which rays pass undeviated for paraxial rays.
Focal points and sign conventions
Each lens has two focal points, one on either side, equidistant from the optical centre for thin symmetric lenses. The focal length f is positive for converging (convex) lenses and negative for diverging (concave) lenses (sign conventions depend on the textbook or teacher; use the one specified in class consistently). Rays coming from infinity parallel to the principal axis converge to the focal point of a converging lens and appear to diverge from the focal point of a diverging lens.
Lens formula derivation idea (qualitative)
The lens formula 1/f = 1/v - 1/u for thin lenses links object distance u, image distance v and focal length f. It can be derived by applying Snell’s law or small-angle approximations at the two refracting surfaces and using geometry for paraxial rays. For Class 10, use the final formula with care about signs and interpret results: positive v means real image on the opposite side of the lens from the object, negative v means virtual image on the same side as the object.
Principal rays for construction
To locate images, three rays are convenient for a convex lens: (1) a ray parallel to the axis refracts through the far focal point; (2) a ray passing through the optical centre goes straight without deviation; (3) a ray going through the near focal point emerges parallel to the axis. For concave lenses, the ray from the object directed parallel to the axis appears to diverge from the near focal point after refraction, and rays must be extended backward to find the virtual image.
Working cases and image characteristics
Typical object positions and resulting images for a convex lens: object beyond 2F gives real, inverted, diminished image between F and 2F; object at 2F gives real, inverted image at 2F same size; object between F and 2F gives real, inverted enlarged image beyond 2F; object at F produces rays parallel and image at infinity; object between lens and F produces virtual, erect, enlarged image on same side as object. Concise diagrams with distances labelled help solve numerical problems using lens formula and magnification relation.
Magnification and image size
Magnification m = image height/object height = v/u (with sign indicating orientation). If m is negative, the image is inverted. Combine this with lens formula to compute image size from object size. In experiments measure u and v carefully and calculate f from multiple trials to reduce error. Lenses are widely used in spectacles, cameras, microscopes and telescopes and this basic theory is applied directly in those devices.
- Convex lens forming a real inverted image at v when object is beyond 2F; compute v using lens formula.
- Concave lens always forming a virtual erect diminished image; locate using ray extensions.
- Using magnification m = v/u to find image height given object height and distances.
- Lens formula: 1/f = 1/v - 1/u
- Magnification m = v / u = height of image / height of object
Combination of Lenses and Lens Power
Why combine lenses?
Combining lenses allows designers to achieve focal lengths and optical powers not possible with a single element, correct aberrations, or build instruments such as microscopes and telescopes where an objective and eyepiece together produce required magnification. In experiments and instruments, two or more thin lenses may be placed either in contact or separated by a small distance along the same principal axis.
Lenses in contact: resultant power
For thin lenses placed in contact (negligible separation), the combined power P is the algebraic sum of individual powers: P = P1 + P2 where Pi = 1/fi (f in metres) and sign conventions apply (convex +, concave -). The resultant focal length f_res is then 1/f_res = 1/f1 + 1/f2. This simple addition is very useful in optical prescriptions and in simple instrument calculations.
Lenses separated: stepwise image formation
When lenses are separated by a finite distance, the image formed by the first lens acts as the object for the second. To find the final image position, calculate the first image using lens formula for lens 1; use that image distance relative to lens 2 (taking sign and separation into account) as the object distance for lens 2; then use lens formula again. Ray tracing step-by-step also gives quick visual understanding of how images move and whether final image is real or virtual.
Power and units
Power P of a lens is defined as P = 1/f with focal length f in metres; power is measured in dioptres (D). A lens of focal length 0.50 m has power +2.0 D. Powers add algebraically for thin lenses in contact, which makes it easy to calculate spectacle prescriptions and combined lens effects when designing simple optical systems.
Practical examples and adjustments
In correcting vision a diverging lens of appropriate negative power compensates for myopia; for hypermetropia a converging lens with required positive power is used. In instruments like microscopes the objective provides high power (short focal length) while the eyepiece provides additional magnification — combined power and geometry set the final magnification. Always convert cm to metres before calculating power and keep sign conventions consistent.
Limitations and approximations
The contact addition P = P1 + P2 holds well when lens thicknesses are small and separation is negligible. For precise optical design, lens thickness, surface curvatures and spacing need careful consideration. For Class 10 and most board questions the thin-lens and contact approximations are sufficient and commonly used in numerical problems and eyewear calculations.
- Compute combined power of lenses of focal lengths +50 cm and +25 cm when in contact.
- Find focal length of a lens having power -2 D.
- Explain how adding a concave lens of negative power with a convex lens reduces overall power.
- Power P = 1 / f (f in metres)
- Combined power in contact: P = P1 + P2
Dispersion of Light and Prisms
Why white light splits
White light is a mixture of many wavelengths (colours). When white light enters a medium like glass, the refractive index varies slightly with wavelength — commonly called dispersion. Shorter wavelengths (violet, blue) are refracted more strongly than longer wavelengths (red). A prism refracts each wavelength differently at its two surfaces; the combined effect produces spatial separation of colours called a spectrum.
How a prism disperses light
When a narrow beam of white light strikes the first face of a triangular prism, different colours are refracted by different angles according to Snell’s law. Inside the prism each component travels along a different path and on emerging at the second surface the components separate further. The emergent beam spreads into the familiar band of colours from red (least deviated) to violet (most deviated).
Deviation and angular separation
The angle through which light is deviated by a prism depends on the prism angle and the refractive index for each wavelength. For small prism angles and paraxial rays, deviation is approximately proportional to (n - 1), so materials with greater dispersion produce larger angular spreads. The phenomenon is reversible: sending the spectrum through an inverted prism recombines the colours back into white light, demonstrating that dispersion itself does not destroy colours — it only separates them.
Applications and implications
Dispersion explains natural rainbows where water droplets act as tiny prisms combined with internal reflection to separate sunlight into colours. In optics dispersion causes chromatic aberration in simple lenses because different colours focus at different distances; designers use achromatic doublets to reduce this effect. Spectrometers use prisms or diffraction gratings to analyse the spectral composition of light for scientific and industrial applications.
Class experiments and observations
In a darkened room, shine a narrow white light through a glass prism onto a screen and observe the spectrum. Measure the deviation angles for different colours roughly by marking positions on the screen. Try recombining the spectrum with a second prism oriented oppositely to restore white light. These exercises make the wavelength-dependence of refractive index and the idea of reversible dispersion clear.
Limits and further ideas
Dispersion depends on material; some glasses have low dispersion and are used to reduce chromatic effects. Quantitative dispersion relations (Cauchy’s formula etc.) are beyond Class 10, but the essential understanding — that refractive index varies with wavelength and causes spectral separation — is central and sufficient for board problems and applications in optics.
- Explain why violet light deviates more than red when passing through a glass prism.
- Sketch rays of white light through a prism showing emergent spectrum order from red to violet.
- Describe recombination of spectrum by a second inverted prism to obtain white light.
- Angle of deviation and prism relations (qualitative at Class 10)
Scattering of Light and Sky Colour
What is scattering?
Scattering is the process by which small particles or molecules redirect light in many directions. In the atmosphere sunlight meets molecules of nitrogen and oxygen and various aerosols and dust particles; some of the incident light energy is scattered away from its original direction. The amount and direction of scattering depends on the size of the particles relative to the wavelength of light.
Rayleigh scattering and the blue sky
When particles are much smaller than the wavelength of light (molecular scale), scattering intensity varies strongly with wavelength — approximately inversely with the fourth power of wavelength (qualitative idea). This means blue light (shorter wavelength) is scattered much more than red. Because of human eye sensitivity and atmospheric composition, the scattered light we see from all directions during the day is predominantly blue, hence the sky looks blue.
Sunrise and sunset colours
At sunrise and sunset sunlight travels a much longer path through the atmosphere to reach the observer. Along this longer path most of the blue and green components have been scattered out of the direct beam, leaving the longer red and orange wavelengths; thus the Sun appears red and the surrounding sky near the horizon takes on warm colours. Aerosols and pollution can enhance reddening by additional scattering and absorption.
Mie scattering and clouds
When particles are comparable in size to the wavelength (e.g., water droplets in clouds), scattering is less wavelength-dependent and tends to scatter all visible wavelengths fairly equally. This produces white appearances — which is why clouds look white from below when sunlight illuminates many droplets equally. Hazy conditions with larger particles scatter differently and can produce whitening or brownish tints depending on particle composition.
Other consequences and observations
Scattering causes distant mountains to appear bluish because light from them is scattered by intervening atmosphere; it limits visibility on polluted days. Scattering is also a consideration in optical communications through the atmosphere and in remote sensing, where the amount and colour of scattered light give information about particle sizes and composition.
Simple classroom demonstration
Create a dilute milk-in-water solution in a glass and shine a narrow beam through it. Observe the scattered light at right angles appearing blue while the transmitted beam appears reddish on the far side as short wavelengths are preferentially scattered. This demonstration mimics Rayleigh scattering qualitatively and helps visualise why the sky is blue and sunsets are red.
- Explain why sunsets are red using the idea of scattering removing short wavelengths.
- Why are clouds white while the sky is blue? Use particle size argument.
- Model experiment using milk in water showing blue scattered light at right angles.
Optical Instruments: Simple Microscope
Purpose and basic design
A compound microscope magnifies tiny objects that are too small to see clearly with the naked eye. It uses two lenses in sequence: a short-focal-length objective lens close to the object that produces a real enlarged intermediate image, and an eyepiece (ocular) that acts as a simple magnifier to produce a further enlarged virtual image for the eye. The tube length between objective and eyepiece and the focal lengths determine the overall magnification and working distances.
How the objective works
The objective has a very short focal length and is placed just a little further from the object than its focal length. It forms a real, inverted and enlarged image of the object at a point inside the tube. This intermediate image is typically located just within the focal length of the eyepiece so that the eyepiece can produce a clear final virtual image at a comfortable viewing distance (often at infinity for relaxed eye).
Eyepiece action and angular magnification
The eyepiece acts like a magnifier: it produces a virtual image of the intermediate real image and allows the eye to view that image with larger angular size. The angular magnification of the eyepiece is approximately 25 cm / f_e (for normal near point of 25 cm) or simply 1 + (25 cm / f_e) depending on whether final image is at near point or at infinity; for Class 10 a qualitative understanding and the product rule for magnifications is sufficient.
Total magnification
Approximately, total magnification M_total = magnification by objective × magnification by eyepiece. The objective provides linear magnification (ratio of image size to object size) while the eyepiece provides angular magnification; the product gives how much larger the final virtual image appears compared to the object as seen by the unaided eye. Practical microscopes specify objective and eyepiece powers, and the total magnification is the product.
Design trade-offs and care
High magnification requires short focal length objectives and precise mechanical adjustments; however resolution (ability to see fine details) depends on lens quality and illumination, not only magnification. Chromatic and spherical aberrations must be corrected in good objectives. Care of a microscope includes cleaning lenses with lens tissue, avoiding contact of slides with objective, using correct illumination and storing covered to prevent dust.
Simple ray diagram
Draw a principal axis, place the objective close to the object, show rays forming a real intermediate image, then draw the eyepiece that takes that image and produces a virtual image at infinity. Label the objective focal length and eyepiece focal length, and indicate the final virtual enlarged image. Understanding these ray paths helps answer questions about image orientation, magnification and focusing adjustments in microscopes.
- Explain qualitatively how a compound microscope magnifies tiny objects using an objective and eyepiece.
- Sketch the path of rays in a microscope showing intermediate real image and final virtual image at infinity.
- Show how changing the eyepiece focal length changes overall magnification.
Optical Instruments: Telescope
Purpose and general types
A telescope collects light from distant objects and produces images that are easier to study. The two main types are refracting telescopes, which use lenses, and reflecting telescopes, which use mirrors. For Class 10 focus on the simple refracting telescope working in normal adjustment where the final image is at infinity, because it is straightforward to analyse and appears in many examination questions.
Basic refracting telescope arrangement
In a basic refracting telescope, a large-aperture objective lens collects parallel rays from a distant object and forms a real image at its focal plane. The eyepiece, a short-focal-length lens, is placed so that it uses this real image as an object and produces parallel rays leaving the eyepiece, creating a final virtual image at infinity. This arrangement reduces strain on the observer’s eye since the eye is relaxed when viewing parallel rays.
Angular magnification
The angular magnification (or magnifying power) of a telescope in normal adjustment is given approximately by the ratio of focal lengths of the objective and the eyepiece: M ≈ f_o / f_e. A long-focus objective and a short-focus eyepiece give high magnification. Angular magnification means the angle subtended by the image at the eye when using the telescope is larger than the angle subtended by the object when seen with the naked eye.
Field of view, brightness and resolution
The objective diameter determines light-gathering power and resolution: larger diameter collects more light and can resolve finer details. Reflecting telescopes allow larger mirrors and avoid chromatic aberration present in lenses. However, very high magnifications require stable mounting and are limited by atmospheric turbulence; magnification without adequate light or resolution gives dim, blurred images.
Design choices and practicalities
Refractors suffer from chromatic aberration unless anti-chromatic designs are used, and manufacturing large quality lenses is difficult and expensive. Reflectors use mirrors and avoid chromatic dispersion but require careful alignment (collimation). Telescopes are mounted on tripods or equatorial mounts to track objects. In classroom problems, apply M = f_o / f_e, draw ray diagrams showing parallel incoming rays focused by objective and emerging as parallel rays from eyepiece, and note that tube length ≈ f_o + f_e in normal adjustment.
Applications
Telescopes are used in astronomy and in terrestrial long-distance observing. Understanding their optics helps explain how choices of objective and eyepiece focal lengths affect magnification, brightness and practical use in observation sessions.
- Calculate magnifying power of a telescope with objective focal length 1000 cm and eyepiece 10 cm: M = 1000/10 = 100x.
- Sketch a refracting telescope in normal adjustment showing lens separations.
- Explain why larger objective diameter helps see fainter stars.
- Approximate magnification of telescope M = f_o / f_e
Human Eye: Structure and Image Formation
Parts of the eye important for optics
The human eye is an optical system. Light enters through the cornea, which provides most of the eye’s refractive power. The aqueous humour, lens and vitreous humour complete the refracting media. The crystalline lens is flexible and changes shape to focus at different distances — this process is called accommodation. The iris controls pupil size and therefore the amount of light reaching the retina. The retina is a light-sensitive screen made of photoreceptor cells (rods and cones) where a real inverted image forms and is converted into nerve signals sent to the brain.
How the eye forms images
For distant objects the ciliary muscles relax and the lens becomes thinner, reducing its optical power so that parallel rays focus on the retina. For near objects the muscles contract, making the lens thicker and more curved so that its focal length decreases and the image again falls on the retina. A normal eye can adjust (accommodate) from the near point (about 25 cm for a young eye) to infinity.
Image characteristics on the retina
The image formed on the retina is real and inverted; however, the brain interprets this signal and perceives objects as upright. The size of the retinal image depends on the object distance and the eye’s focal length. Sharpness depends on correct focusing and the quality of the refracting surfaces, while sensitivity to colour comes from cone cells that respond to different wavelengths.
Accommodation and limits
Accommodation range depends on lens flexibility and decreases with age; the closest distance that can be seen clearly is the near point, and the farthest distance, the far point, is normally at infinity. Problems with the cornea, lens or eyeball shape cause refractive errors such as myopia and hypermetropia. Regular eye tests determine corrective lenses needed to restore clear focus on the retina.
Care and adaptation
Eyes adapt to a wide range of lighting by changing pupil size and through photochemical processes in rods and cones. Protect eyes from strong sunlight and bright lasers; avoid prolonged strain and ensure regular breaks during near work. Simple diagrams showing rays entering the eye and focusing on the retina, and sketches of lens curvature during accommodation, help visualise how the eye works and why certain defects arise.
- Draw a simple ray diagram showing light from a distant object focused on the retina by cornea and lens.
- Explain accommodation when viewing a near object and how lens curvature changes.
- Describe why the image on the retina is inverted but we perceive it upright.
Defects of Vision and Their Correction
Common refractive errors
Three common defects are myopia (short-sightedness), hypermetropia (long-sightedness) and presbyopia (age-related loss of accommodation). Myopia occurs when the eye focuses parallel rays in front of the retina because the eye is too long or the refracting power is too strong; distant objects appear blurred. Hypermetropia occurs when the eye focuses rays behind the retina because the eye is too short or lens power is insufficient; near objects are blurred. Presbyopia results from loss of elasticity of the crystalline lens with age, reducing accommodation and making near vision difficult.
Astigmatism and other defects
Astigmatism arises when the cornea or lens has different curvatures in different meridians, so rays in one plane focus at a different point than rays in another plane. This causes distorted or blurred vision for all distances. Other issues include cataract (clouding of the lens) and colour blindness (difficulty distinguishing particular colours due to photoreceptor deficiencies), which have different treatment approaches.
Spectacle correction principles
Myopia is corrected using a concave (diverging) lens of appropriate negative power. This lens forms a virtual image of a distant object at the person’s far point so the eye can focus it on the retina. Hypermetropia is corrected using a convex (converging) lens of suitable positive power that enables the near object to be focused on the retina. Presbyopia may require bifocal lenses that combine two powers: one for distance and one for near tasks.
Calculating corrective power
To find the power of spectacle lens, use the lens formula 1/f = 1/v - 1/u with correct sign conventions. For distant vision correction typically the object is at infinity (u = ∞), simplifying calculations; for near vision corrections use u = -25 cm or the required near distance and choose v equal to the person's far point or the image position that allows comfortable viewing. Then compute power P = 1/f (in metres) and express in dioptres (D). Remember signs: diverging lenses have negative focal length and negative power, converging lenses positive.
Contact lenses and surgery
Contact lenses sit on the cornea and function like spectacles but with less distortion for some activities. Surgical options like LASIK alter corneal curvature to change focusing power permanently; these are medical procedures requiring specialist assessment. For school-level problems, focus on spectacles and simple calculations, and always include clear ray diagrams showing how the corrective lens moves the image onto the retina.
- A myopic eye has its far point at 80 cm. Find the power of glasses required to see distant objects clearly (object at infinity).
- Explain how a convex lens helps a hypermetropic person to read a book at 25 cm.
- Describe presbyopia and why bifocal lenses are sometimes used.
- Lens formula to compute corrective lens power: 1/f = 1/v - 1/u
- Power P = 1/f (in metres)
Optical Fibres and Communication
Principle of operation
Optical fibres guide light along long distances using repeated total internal reflection (TIR). A typical fibre has a central core of glass or plastic with refractive index n_core surrounded by cladding with a slightly lower refractive index n_clad. Light entering the core within a certain acceptance angle is internally reflected at the core‑cladding boundary and remains trapped, travelling along the fibre with low loss. This principle allows the transmission of pulses of light representing digital information with high bandwidth and minimal interference.
Acceptance cone and numerical aperture
Rays that enter within a certain cone of angles (relative to the fibre axis) will meet the core‑cladding interface at angles exceeding the critical angle and therefore be totally internally reflected. The numerical aperture (NA) quantifies how large this acceptance cone is. For a step-index fibre, NA = sqrt(n_core^2 - n_clad^2). A larger NA means the fibre accepts light from a wider range of input angles, making coupling easier but potentially increasing modal dispersion in multimode fibres.
Types of fibres and losses
Fibres may be single-mode (very thin core, allowing one propagation mode) or multimode (thicker core supporting many modes). Single‑mode fibres have low modal dispersion and are used for long-distance communication with lasers; multimode fibres are used for short distances and simpler equipment. Losses arise from absorption by the glass material, scattering due to microscopic inhomogeneities, and imperfect reflections at bends or connectors. High-quality manufacturing and proper cladding minimise these losses.
Applications in communication and medicine
Optical fibres form the backbone of modern telecommunications because they provide very high data rates, are immune to electromagnetic interference, and allow secure transmission. They are used in backbone internet cables, local networks and cable TV. In medicine flexible fibres are used in endoscopes to relay images from inside the body, enabling minimally invasive diagnosis and surgery. Fibres are also used in sensors, illumination and lighting displays.
Practical demonstration and handling
In demonstrations, a light source coupled to a fibre shows that light follows bends without escaping if bends are gentle and the TIR condition holds. Sharp bends cause leakage and signal loss. When designing fibre systems, ensure proper connectors, avoid sharp bends, and protect fibres from mechanical damage. For Class 10, understanding the TIR principle and basic NA relation is sufficient to explain why fibres work and why they are important in communication.
- Explain why cladding is necessary in an optical fibre and how it helps total internal reflection.
- Describe qualitatively how digital signals are sent through optical fibres as light pulses.
- Show how bending a fibre too sharply can cause loss of light due to failure of TIR.
- Numerical aperture NA = sqrt(n_core^2 - n_clad^2) (qualitative use at Class 10)
Formation of Colours and Colour Addition
What determines colour?
Colour depends on which wavelengths of visible light enter the eye. Objects appear coloured because they reflect, transmit or emit particular wavelengths and absorb others. White light contains many wavelengths and can be separated into colours by dispersion. Filters and pigments work by selectively absorbing certain wavelengths so that the transmitted or reflected light determines perceived colour.
Additive colour mixing
Additive mixing refers to combining coloured lights. The primary colours of light are red, green and blue. When these lights overlap on a screen they add their intensities: red + green = yellow; green + blue = cyan; blue + red = magenta; and red + green + blue in the right proportions produces white light. This principle underlies displays (TVs, monitors) that use tiny red, green and blue elements to create images of many colours by controlling intensities.
Subtractive colour mixing
Subtractive mixing is relevant for paints, dyes and inks that absorb some wavelengths and reflect or transmit others. The primary subtractive colours are cyan, magenta and yellow. Combining cyan and yellow absorbs red and blue leaving green, etc. In printing, combinations of cyan, magenta and yellow inks produce a wide range of colours; black ink is often added (CMYK) to improve depth and contrast.
Perception and physiology
The human eye has three types of cone cells sensitive to different wavelength ranges roughly corresponding to red, green and blue. The brain combines signals from these cones to produce the sensation of colour. Colour blindness often arises when one type of cone is missing or defective, commonly causing difficulty distinguishing red and green colours.
Practical demonstrations
Use three coloured LEDs (red, green, blue) to project onto a white screen and show additive mixing. For subtractive mixing overlap transparent coloured sheets to show how colours change. Use a prism to produce a spectrum and then recombine colours using filters or a second prism to show that white light is a mixture of colours. These hands-on activities clarify how both additive and subtractive systems create perceived colours.
Applications
Understanding colour addition and subtraction is essential in art, printing, photography and digital displays. It also connects to dispersion and chromatic effects in optics: lenses and instruments must manage colour separation to produce sharp images in colour systems.
- Demonstrate additive mixing: overlap red and green lights to produce yellow on a screen.
- Show subtractive mixing using transparent coloured sheets (cyan + yellow ≈ green).
- Explain why a white shirt appears coloured under a coloured light source.
Chromatic Aberration and Achromatic Lenses
Origin of chromatic aberration
Chromatic aberration occurs because a lens has slightly different refractive indices for different wavelengths of light — a property called dispersion. Shorter wavelengths (blue) are refracted more strongly than longer wavelengths (red), so they focus at different points along the optical axis. As a result, a simple single lens will not bring all colours into the same focus, producing coloured fringes around high-contrast edges and reducing sharpness of images.
Types and effects
There are two kinds: longitudinal chromatic aberration where different colours focus at different distances from the lens, and lateral chromatic aberration where magnification varies with wavelength causing colour fringing towards the edges of the image. In photography and microscopy these aberrations degrade image quality, particularly at high magnifications and with wide apertures.
How achromatic lenses work
An achromatic doublet pairs two lens elements made from glasses with different dispersion properties — typically a positive (convex) crown-glass element with low dispersion and a negative (concave) flint-glass element with higher dispersion. By choosing curvatures and materials appropriately the chromatic dispersion of one lens is made to cancel that of the other for two selected wavelengths (often red and blue), bringing them close to the same focus and greatly reducing visible chromatic aberration for white light.
Practical design considerations
Achromatic doublets greatly improve image sharpness and are widely used in microscopes, telescopes, camera objectives and eyepieces. Further correction (apochromatic lenses) addresses three wavelengths and reduces residual colour further. Coatings and correct lens spacing also help control other aberrations like spherical aberration, coma and astigmatism alongside chromatic issues.
Class-level understanding
Students should be able to explain why chromatic aberration arises (refractive index depends on wavelength), describe its visual effects and state that achromatic doublets reduce this by combining two types of glass. Detailed optical design equations are beyond Class 10, but the qualitative mechanism and practical significance are essential for understanding how high-quality optical instruments produce clear colour images.
Simple observation
Using a magnifying glass on a high-contrast edge under white light often reveals coloured fringes; replacing it with an achromatic lens or viewing in monochromatic light reduces or removes the fringes. This simple activity demonstrates the chromatic dispersion that causes chromatic aberration and how achromatic design mitigates it.
- Explain why a simple magnifying glass shows coloured fringes around high-contrast edges when using white light.
- Describe how an achromatic doublet reduces chromatic aberration compared to a single lens.
- Suggest practical ways instrument designers reduce chromatic effects in cameras or microscopes.
Reflection and Refraction Experiments and Instrumentation
Importance of experiments
Practical experiments in reflection and refraction transform abstract laws into observable facts. They build skills in measurement, ray-tracing, drawing neat diagrams and analysing errors. Key experiments in the light unit include verifying the laws of reflection, measuring refractive index using Snell’s law, determining the critical angle, verifying lens and mirror formulae and observing dispersion by a prism.
Typical apparatus and set-up
Common equipment includes a ray box with slits, pins for locating object points, plane and spherical mirrors, glass or perspex slabs, semi-circular blocks, convex and concave lenses, optical bench or ruler for alignment, and a protractor for angle measurements. For refraction experiments a rectangular block or semi-circular block is placed on paper, incident and emergent rays are traced, normals drawn at the points of incidence and angles measured accurately with a protractor.
Verifying Snell’s law
Using a glass slab or semi-circular block, vary the angle of incidence i and measure the angle of refraction r. Tabulate sin i and sin r and plot sin i on the vertical axis against sin r on the horizontal axis; the plot should be a straight line passing through the origin. The slope of this line gives the refractive index ratio n2/n1. Repeat measurements for accuracy and discuss possible sources of error like alignment, thickness of the ray and parallax when reading the protractor.
Determining focal length and mirror formula
To find the focal length of a convex lens, use the distant object method (object at infinity) and focus the image on a screen; measure v as the focal length. Alternatively use the lens formula by placing the object at various distances u, measuring v and computing f from 1/f = 1/v - 1/u in each case and averaging. For mirrors use ray boxes and screens to locate images and confirm mirror formula. Note precision in locating sharp images and avoid parallax errors when measuring distances.
Measuring critical angle
A semi-circular block allows rays to enter through the flat face and strike the curved face at varying angles. By changing the incident position and observing when the refracted ray just disappears at the curved face, students can approximate the critical angle. Use Snell’s law with r = 90° to compute refractive index and compare with tabulated values.
Recording results and error analysis
Present data in clear tables with units, calculate averages and compare with expected values. Discuss systematic errors (e.g., imperfect apparatus, misalignment) and random errors (reading variations). Suggest improvements such as using finer slits, better alignment, repeating trials and using digital equipment if available. These practices prepare students for practical exams and give a realistic sense of experimental physics.
- Design an experiment to verify Snell's law using a semi-circular block and plot sin i against sin r to find refractive index.
- Describe a procedure to find focal length of a convex lens using distant object method.
- Explain common sources of error in ray-tracing experiments and how to reduce them.
Key Concepts
- Ray
- A line in the direction of propagation of light, showing its path in geometric optics.
- Wavefront
- A surface connecting points of a light wave that have the same phase.
- Reflection
- The bouncing back of light from a surface into the same medium.
- Refraction
- The bending of light when it passes from one medium to another due to speed change.
- Refractive index
- The ratio of speed of light in vacuum to speed in a medium, indicating optical density.
- Critical angle
- The angle of incidence in the denser medium beyond which total internal reflection occurs.
- Total internal reflection
- Complete reflection of light back into the denser medium when incidence exceeds critical angle.
- Focal length
- The distance from the optical centre of a lens or from the pole of a mirror to its principal focus.
- Lens formula
- The relation 1/f = 1/v - 1/u that connects focal length, object distance and image distance.
- Magnification
- The ratio of image height to object height, also equal to image distance divided by object distance for thin lenses.
- Dispersion
- Separation of white light into constituent colours due to wavelength-dependent refraction.
- Chromatic aberration
- Colour fringing in images caused by different focal points for different wavelengths.
- Optical fibre
- A thin transparent fibre that transmits light by repeated total internal reflection.
- Near point
- The closest distance at which the eye can focus an object comfortably, typically 25 cm for a normal eye.
- Power of lens
- The reciprocal of focal length measured in metres, expressed in dioptres (P = 1/f).
Practice Questions
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State the two laws of reflection. / परावर्तन के दो नियम बताइए।
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The incident ray, the reflected ray and the normal at the point of incidence lie in the same plane; and the angle of incidence equals the angle of reflection. / परावर्तन के बिंदु पर आगमन किरण, परावर्तित किरण और सामान्य एक ही तल में होते हैं; और पदार्पण कोण परावर्तन कोण के बराबर होता है।
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A ray of light in air strikes a glass surface (n = 1.5) at an angle of 30°. Find the angle of refraction. / हवा में एक प्रकाश किरण 30° पर कांच की सतह (n = 1.5) से टकराती है। अपवर्तन कोण ज्ञात कीजिए।
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Using Snell's law: n1 sin i = n2 sin r so 1.0 × sin30° = 1.5 × sin r ⇒ 0.5 = 1.5 sin r ⇒ sin r = 1/3 ⇒ r ≈ 19.47°. / स्नेल के नियम से: 1.0 × sin30° = 1.5 × sin r ⇒ 0.5 = 1.5 sin r ⇒ sin r = 1/3 ⇒ r ≈ 19.47°।
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An object 5 cm tall is placed 30 cm from a convex lens of focal length 10 cm. Find the image distance and image size. / एक 5 सेमी ऊँचा वस्तु 30 सेमी दूरी पर एक उत्तल लेंस (f = 10 सेमी) से रखा गया है। छवि दूरी और छवि आकार निकालिए।
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Lens formula: 1/f = 1/v - 1/u ⇒ 1/10 = 1/v - 1/(-30) [u = -30 cm by sign convention] ⇒ 1/10 = 1/v + 1/30 ⇒ 1/v = 1/10 - 1/30 = (3 -1)/30 = 2/30 = 1/15 ⇒ v = 15 cm. Magnification m = v/u = 15/(-30) = -1/2 so image height = m × object height = -1/2 × 5 cm = -2.5 cm (negative means inverted). / लेंस सूत्र: 1/10 = 1/v - 1/u ⇒ 1/v = 1/10 - 1/30 = 1/15 ⇒ v = 15 सेमी। आवर्ध m = v/u = 15/(-30) = -1/2 ⇒ छवि ऊँचाई = -2.5 सेमी (नकारात्मक का अर्थ उल्टी)।
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Define critical angle and calculate it for a glass-air interface when n_glass = 1.5. / क्रिटिकल कोण परिभाषित करें और कांच‑हवा सीमा के लिए इसे निकालिए यदि n_कांच = 1.5।
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Critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°; sin c = n_rarer / n_denser. For glass to air: sin c = 1.0 / 1.5 ⇒ c = arcsin(0.6667) ≈ 41.8°. / क्रिटिकल कोण वह पदार्पण कोण है जिसके लिए पतले माध्यम में अपवर्तन कोण 90° होता है; sin c = n_दुर्लभ / n_घन। कांच‑हवा के लिए sin c = 1/1.5 ⇒ c ≈ 41.8°।
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Why does the sky appear blue during the day and red at sunrise/sunset? / दिन के समय आकाश नीला और सूर्योदय/सूर्यास्त पर लाल क्यों दिखता है?
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Blue light is scattered more strongly by air molecules (Rayleigh scattering) because it has shorter wavelength, so scattered light from the sky appears blue. At sunrise/sunset sunlight passes through more atmosphere, removing shorter wavelengths by scattering and leaving longer wavelengths (red/orange), so Sun and sky near horizon look red. / नीली रोशनी छोटी तरंगदैর্ঘ्य होने के कारण वायु अणुओं द्वारा अधिक बिखरी जाती है (रेलेघ बिखराव), इसलिए आकाश नीला दिखता है। सूर्योदय/सूर्यास्त में प्रकाश अधिक वायुमंडल से होकर गुज़रता है, जिससे छोटी तरंगें बिखर जाती हैं और लाल-नारंगी तरंगें बचती हैं, इसीलिये क्षितिज के पास लाल दिखता है।
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Describe with a ray diagram how a concave mirror forms a real image when the object is beyond the centre of curvature. / एक अवतल (concave) दर्पण तब वास्तविक छवि कैसे बनाता है जब वस्तु वक्रता के केंद्र से बाहर हो, बताइए और किरण रेखाचित्र दीजिए।
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For an object beyond the centre C, draw principal axis and locate P (pole), C and F. From a point on the object draw one ray parallel to the axis which after reflection passes through F, and another ray towards C which reflects back on itself. The reflected rays meet in front of the mirror between F and C forming a real, inverted and diminished image. / वस्तु C से बाहर होने पर प्रमुख अक्ष पर P, C और F दर्शाइए। वस्तु बिंदु से एक किरण धुरी के समानांतर खींचें जो परावर्तन के बाद F से गुज़रेगी, और एक किरण C की ओर भेजें जो उस पर वापस परावर्तित होगी। परावर्तित किरणें मिलकर दर्पण के सामने F और C के बीच एक वास्तविक, उल्टी और छोटी छवि बनाती हैं।
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How do optical fibres use total internal reflection to transmit signals? / ऑप्टिकल फाइबर संचार के लिए टोटल इंटरनल रिफ्लेक्शन का उपयोग कैसे करते हैं?
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Optical fibres have a core with higher refractive index and cladding with lower index. Light launched within the acceptance angle strikes the core-cladding boundary at angles greater than critical angle, so it undergoes total internal reflection and remains guided along the core with minimal loss, allowing long-distance signal transmission as light pulses. / ऑप्टिकल फाइबर के कोर का अपवर्तनांक क्लैडिंग से अधिक होता है। जब प्रकाश स्वीकार्य कोण के भीतर फाइबर में प्रवेश करता है, तो वह कोर‑क्लैडिंग सीमा पर क्रिटिकल कोण से अधिक कोण पर गिरता है और पूर्ण आंतरिक परावर्तन करता है, जिससे प्रकाश कोर के भीतर मार्गदर्शित होकर कम हानि पर लंबी दूरी तय करता है।
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A person cannot see objects closer than 150 cm clearly. What defect is this and what lens is needed for correction? Find its power. / एक व्यक्ति 150 सेमी से निकट की वस्तुएँ साफ़ नहीं देख पाता। यह दोष क्या है और किस प्रकार का लेंस चाहिए? उसकी शक्ति ज्ञात कीजिए।
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This is hypermetropia (long-sightedness) because near point is farther than normal (25 cm). To read at 25 cm, a convex lens is needed to produce a virtual image at 150 cm when object is at 25 cm. Using lens formula 1/f = 1/v - 1/u with sign convention (object u = -25 cm, image v = -150 cm since virtual image on same side): 1/f = 1/(-150) - 1/(-25) = -1/150 + 1/25 = (-1 +6)/150 = 5/150 = 1/30 ⇒ f = 30 cm ⇒ power P = 1/f (in metres) = 1/0.30 = +3.33 D (approx +3.3 D). / यह दूरदृष्टि (हाइपरमेट्रोपिया) है क्योंकि निकट बिंदु सामान्य (25 सेमी) से दूर है। समीकरण से 1/f = 1/v - 1/u लेते हैं (u = -25 सेमी, v = -150 सेमी) ⇒ 1/f = -1/150 + 1/25 = 1/30 ⇒ f = 30 सेमी ⇒ शक्ति P = 1/0.30 ≈ +3.33 D।
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Explain dispersion and how two prisms can recombine a spectrum back into white light. / डिस्पर्शन क्या है और कैसे दो प्रिज्म मिलाकर स्पेक्ट्रम को पुनः श्वेत प्रकाश में बदल सकते हैं?
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Dispersion is separation of white light into component colours because refractive index depends on wavelength, so different colours refract by different angles in a prism. If a second identical prism is placed inverted relative to the first, the angular separations caused by the first prism are reversed by the second, causing the colours to recombine and restore white light. This shows dispersion is a reversible process. / डिस्पर्शन वह प्रक्रिया है जिसमें श्वेत प्रकाश के विभिन्न तरंगदैর্ঘ्यों के अलग‑अलग अपवर्तन के कारण रंग अलग हो जाते हैं। यदि दूसरा समरूपी प्रिज्म पहले के उल्टे क्रम में रखा जाए तो वह पहले द्वारा विभाजित रंगों को फिर से उलट कर उनका संलयन कर देता है और श्वेत प्रकाश बहाल हो जाता है, जो दर्शाता है कि डिस्पर्शन प्रतिवर्ती है।
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Write the mirror formula and magnification formula. State the sign convention used in problems. / दर्पण सूत्र और आवर्धन सूत्र लिखिए। समस्याओं में प्रयुक्त संकेत परिभाषा बताइए।
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Mirror formula: 1/f = 1/v + 1/u. Magnification m = height of image / height of object = v/u. Sign convention typically: distances measured in direction of incident light are positive for lenses/mirrors depending on the convention taught — commonly object distance u is taken as negative (object in front of mirror), image distance v is positive for real images in front of mirror and negative for virtual images behind mirror, and focal length f is positive for concave mirrors and negative for convex mirrors. Use the specific convention your teacher requires consistently. / दर्पण सूत्र: 1/f = 1/v + 1/u। आवर्धन m = छवि ऊँचाई / वस्तु ऊँचाई = v/u। संकेत नियम: सामान्यतः वस्तु दूरी u सामने की दिशा में नकारात्मक ली जाती है, वास्तविक छवि की दूरी v सामने सकारात्मक और आभासी छवि के लिए नकारात्मक होती है; अवधान f अवतल दर्पण के लिए धनात्मक और उत्तल के लिए ऋणात्मक। जो भी संकेत नियम कक्षा में बताया गया हो उसे निरंतर प्रयोग करें।
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Draw and label a ray diagram for a convex lens when the object is placed at the principal focus. / एक उत्तल लेंस का किरण रेखाचित्र बनाइए और लेबल कीजिए जब वस्तु प्रमुख फोकस पर रखी हो।
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When the object is at the principal focus F of a convex lens, rays after refraction emerge parallel and the image is formed at infinity. In the diagram draw principal axis, lens centre O, object at F on left; draw a ray from top of object parallel to axis which refracts through focal point on right (but here object at F so this ray becomes parallel); draw a ray through centre undeviated. The refracted rays are parallel, so they meet at infinity indicating image at infinity. / यदि वस्तु प्रमुख फोकस F पर है तो उत्तल लेंस के बाद किरणें समांतर निकलती हैं और छवि अनन्त पर बनती है। चित्र में धुरी, लेंस केंद्र O और वस्तु F पर रखें; वस्तु बिंदु से एक किरण धुरी के समानांतर खींचें और एक किरण लेंस के केंद्र से गुजरती हुई बिना विचलन के निकलती दिखाइए। परावर्तित/अपवर्तित किरणें समांतर होंगी, इसलिये छवि अनन्त पर स्थित है।
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