NDA Light previous year questions with solutions

Answer: (b) electric field only

Method. the force of light on charges in the retina is due to the electric field

The magnetic force on a charge is smaller than the electric force by about v/⁠c, which is tiny for the slow electrons in the retinal molecules, so the eye responds to the electric field of the wave only. The eye is sensitive to visible light, not infrared, so (d) is wrong. Standard result.

Answer: (d) Polarization

Method. only transverse waves can be polarised

Refraction, diffraction and interference occur for longitudinal waves (sound) as well. Polarization, restriction of vibrations to one plane, is possible only for transverse waves, hence it supports the transverse nature of light (standard result).

Answer: (b) half

Method. radiation force = 2IA/⁠c on a perfect reflector at normal incidence

Pressure P = 2I/⁠c, force F = P x A = 2IA/⁠c; with I fixed, halving A halves F. (Standard result, radiation pressure.)

Answer: (c) 2 and 3 only

Method. properties of electromagnetic waves

(1) False: elastic waves need a material medium; EM waves are not mechanical and are transverse. (2) True: they travel in vacuum (TP9-A01, TP9-A02). (3) True: E and B are mutually perpendicular and perpendicular to the direction of travel. (4) As printed, 'meters per second' makes the speed 3 x 10^5 m/⁠s, a thousand times too small (c = 3 x 10^8 m/⁠s = 3 lakh km/⁠s), so (4) is false as printed. Hence 2 and 3 only.

Answer: (b) About 8 minutes

Method. finite speed of light, Sun-Earth distance

t = d/⁠c = 1.5 x 10^11 m / 3 x 10^8 m/⁠s = 500 s = about 8.3 minutes, i.e. about 8 minutes (TP9-A02 gives the finite speed of light).

Answer: (a) Ultraviolet waves

Method. ultraviolet radiation and fluorescence

Security features on currency notes fluoresce under ultraviolet light, so UV lamps detect forged notes; infrared, radio waves and microwaves are not used for this. Standard result (electromagnetic spectrum).

Answer: (a) The image will accelerate towards the centre of curvature.

Method. image position of a concave mirror as the object moves from infinity towards C

For an object beyond C the image lies between F and C. As the object moves from infinity towards C, the image moves from F towards C, so it moves towards the centre of curvature (not past it, and not towards the pole). Its speed is (v/⁠u)² times the object's speed, and both the factor v/⁠u (from 0 to 1) and the object's speed increase, so the image accelerates. Hence (a). Mirror formula, 1/⁠v + 1/⁠u = 1/⁠f.

Answer: (a) The image will remain unchanged.

Method. reflection at plane mirrors does not depend on the medium

The mirrors work by reflection (angle of incidence = angle of reflection), which does not depend on the refractive index of the medium around them; the periscope's inclined mirrors give the same erect image. A completely transparent liquid absorbs nothing, so brightness is unchanged. Hence the image is unchanged (taking the end windows as ideal). Standard result.

Answer: (b) It is a convex mirror.

Method. image formed by a convex mirror

A convex mirror always forms a virtual, erect, diminished image between the pole and the focus whatever the object position. A plane mirror's image is behind it at the same distance, and a concave mirror can give images elsewhere (even beyond the focus, real or enlarged). So the mirror is convex. Standard result.

Answer: (a) between pole (P) and the focus (F), behind the mirror

Method. convex mirror image for any real object

A convex mirror always forms a virtual, erect, diminished image behind the mirror between the pole P and the focus F, for an object anywhere between infinity and P(mirror formula with f positive: v is positive and less than f). Standard result.

Answer: (b) At the centre of curvature

Method. concave mirror: object at C gives image at C

An object at the centre of curvature (u = −⁠R = −⁠2f) gives, by 1/⁠v + 1/⁠u = 1/⁠f with f = −⁠R/⁠2, v = −⁠R: the image is at C itself, real, inverted and the same size. Standard result.

Answer: (c) 0.75 m

Method. plane-mirror image: full-length mirror needs half the person's height

By the law of reflection, rays from head and feet reach the eye by striking the mirror at points half-way between eye and head, and between eye and feet, so the mirror length needed is H/⁠2 = 1.5/⁠2 = 0.75 m. This is independent of her distance from the mirror (standard result).

Answer: (c) 12

Method. mirror formula with magnification m = −⁠v/⁠u for each end, Cartesian sign convention

Convention: distances from the pole, positive along the incident light, so for a concave mirror f is negative. Take f → −⁠f. A at 2f: u = −⁠2f, image at v = −⁠2f, m = −⁠v/⁠u = −⁠1, so A′D′ = h (inverted, real). B at 3f: 1/⁠v = 1/⁠(−⁠f) − 1/⁠(−⁠3f) = −⁠2/⁠(3f), so v = −⁠1.5f; m = −⁠v/⁠u = −⁠(−⁠1.5f)/⁠(−⁠3f) = −⁠0.5, so B′C′ = 0.5h. Ratio B′C′ : A′D′ = 1/⁠2. B′C′ and A′D′ are the transverse heights of the image at B and at A (the rectangle is small compared with f). Verified by calculation.

Answer: (a) α = 0

Method. two mirrors at 90° act as a corner reflector: reflected ray is antiparallel to the incident ray

For mirrors at right angles the total deviation is 180° whatever the angle of incidence: the emergent ray is exactly antiparallel to the incident ray (deviation 2(i + r) with i + r = 90°). Rotating the pair rigidly about the edge Y by θ keeps them at 90° (X′Y ⊥ YZ′, as drawn), so the emergent ray stays antiparallel to the unchanged incident ray, hence in the same direction as before. So α = 0. Standard result for a corner reflector.

Answer: (a) focal length goes to infinity.

Method. mirror formula 1/⁠v + 1/⁠u = 1/⁠f applied to a plane mirror

A plane mirror forms an image at v = −⁠u (same distance behind). Then 1/⁠v + 1/⁠u = 0, so 1/⁠f = 0 and f → ∞. A plane mirror is a spherical mirror of infinite radius of curvature (standard result, R = 2f → ∞).

Answer: (b) erect, virtual and same size.

Method. image formed by a plane mirror

A plane mirror always gives an erect, virtual image, laterally inverted, the same size as the object and as far behind the mirror as the object is in front (standard result, m = +1). Only (b) matches.

Answer: (c) virtual and is laterally inverted.

Method. plane mirror image

The image is virtual (cannot be caught on a screen), erect, the same size, as far behind as the object is in front, and laterally inverted (C9-A05). (b) is wrong: it is not nearer than the object.

Answer: (a) It applies only to spherical mirrors.

Method. the formula 1/⁠f = 1/⁠v + 1/⁠u is the mirror formula

With the New Cartesian Sign Convention the mirror formula is 1/⁠v + 1/⁠u = 1/⁠f (C9-A44), while the lens formula is 1/⁠v − 1/⁠u = 1/⁠f (C9-A89). The formula given is valid only for spherical mirrors; e.g. for a lens with u = -2f this formula would give v = 2f/⁠3 instead of the true v = 2f. So (a).

Answer: (b) R = 2f

Method. R = 2f for a spherical mirror of small aperture

The focus lies midway between pole and centre of curvature, so R = 2f (C9-A16, TP9-A08).

Answer: (c) The image will be formed at infinity

Method. concave mirror, object between F and P

The image is behind the mirror: virtual, erect and enlarged (C9-A24). It is at infinity only when the object is at F (C9-A23), so (c) is not correct.

Answer: (c) The image is virtual, erect and laterally inverted.

Method. properties of a plane mirror image

A plane mirror gives a virtual, erect image of the same size, as far behind as the object is in front, laterally inverted (C9-A05, C9-A98). Its focal length is infinite (R is infinite), not zero or 200 cm, and the size equals the object's (m = +1). Only (c) is right.

Answer: (a) It will never form an inverted image.

Method. images in a convex mirror

A convex mirror always gives a virtual, erect, diminished image (C9-A27, C9-A28), so (a) is correct. (c) is wrong for the same reason. (b): as u goes from infinity toward P the image moves from F toward the pole, away from the focus. (d): at u = -f/⁠2, v = fu/⁠(u − f) = f/⁠3, m = -v/⁠u = 2/⁠3 < 1, still diminished (checked with Fraction).

Answer: (a) Reflection of light

Method. a simple periscope uses two plane mirrors

A simple periscope has two plane mirrors parallel at 45 degrees; it works by reflection of light (C9-A05). Prism periscopes use total internal reflection (TP9-A27), but the question says a simple periscope, so (a).

Answer: (b) convex

Method. convex mirrors as rear-view mirrors

A convex mirror always gives a virtual, erect, diminished image and so a wider field of view (C9-A29), which is why it is the rear-view mirror.

Answer: (b) sends parallel rays

Method. concave mirror with source at the focus gives a parallel beam

The bulb is placed at the focus of the concave reflector, so the reflected rays leave parallel to the axis and give a powerful long-range beam (C9-A25).

Answer: (b) The phenomenon of total internal reflection can occur only if light is passing from a medium of high refractive index to a medium of low refractive index and at an angle greater than the critical angle.

Method. conditions for total internal reflection

TIR requires light going from the denser (higher refractive index) to the rarer medium, and an angle of incidence greater than the critical angle sin c = n_2/⁠n_1. In (a) and (d) the direction is reversed, which is impossible; (c) omits the angle condition (at small angles the ray is partly refracted out). Standard result.

Answer: (c) 13d/⁠12

Method. apparent depth through successive layers: sum of t/⁠n for each layer (observer looking normally)

Convention: normal viewing, apparent depth = real thickness/⁠refractive index for each layer. Water: d/⁠(4/⁠3) = 3d/⁠4. Slab: (d/⁠2)/⁠(3/⁠2) = d/⁠3. Total = 3d/⁠4 + d/⁠3 = 9d/⁠12 + 4d/⁠12 = 13d/⁠12. Verified by calculation.

Answer: (a) 3000 Å to 5000 Å

Method. wavelength in a medium λ_m = λ/⁠n(frequency fixed)

Frequency is unchanged on entering water, while v = c/⁠n, so λ_water = λ_air/⁠n. 4020/⁠1.34 = 3000 Å and 6700/⁠1.34 = 5000 Å. Range 3000 Å to 5000 Å. Verified by calculation.

Answer: (c) π/⁠3

Method. Snell's law across parallel slabs: n sin θ is invariant

For parallel plane boundaries n_1 sin θ_1 = n_2 sin θ_2 at each interface, so the product is invariant and the widths do not matter. Final medium is air: sin θ = 1 × sin(π/⁠3), θ = π/⁠3. Intermediate check: in the ⁠√3⁠ slab sin θ = (⁠√3/⁠2)/⁠√3⁠ = 0.5, so no total internal reflection at any interface. Verified by calculation.

Answer: (b) π/⁠3

Method. normal incidence on BC for retracing; TIR at AB

With n = 1.5, the critical angle is sin^-1(1/⁠1.5) = 41.8°, so a ray at π/⁠3 = 60° to the normal of AB is totally reflected at AB (not refracted out). After reflection it makes 90° − 60° = 30° with AB and must strike BC perpendicularly to retrace its path. In the triangle formed by the ray, AB and BC, the angles are 30° at AB, 90° at BC, so angle at B: θ = 180° − 90° − 30° = 60° = π/⁠3. Check: with θ = 60° the ray normal to BC is directed 30° below AB, consistent with the reflected ray. Standard geometry.

Answer: (b) Refraction of light

Method. apparent depth by refraction

Light from the bottom bends away from the normal on passing from water to air, so it appears to come from a shallower point: apparent depth = real depth/⁠n. This is refraction. Standard result.

Answer: (a) X < Y

Method. speed of light in a medium is v = c/⁠n

In carbon disulfide n ≈ 1.63 > 1, so X = Y/⁠n < Y. Light travels fastest in vacuum. Standard result.

Answer: (a) bending towards the normal

Method. Snell's law: ray entering a denser medium bends towards the normal

Air to water is rarer to denser (n 1 to 1.33): sin i = n sin r gives r < i, so the ray bends towards the normal. With a non-zero angle of incidence it does bend; it is neither straight nor wholly reflected. Standard result.

Answer: (c) (2/⁠3)c

Method. refractive index n = c/⁠v

v = c/⁠n = c/⁠(3/⁠2) = (2/⁠3)c. Standard definition of absolute refractive index.

Answer: (c) refraction of starlight

Method. atmospheric refraction causes twinkling

Starlight passes through layers of air whose density, hence refractive index, keeps changing; the ray bends by a randomly varying amount, so the apparent position and brightness flicker. This is refraction (atmospheric refraction), not dispersion or reflection (standard result).

Answer: (a) refraction

Method. twinkling of stars is atmospheric refraction

Continuous refraction of starlight through the changing atmosphere makes the star seem to flicker (C10-A51, C10-A53).

Answer: (c) Light speeds down as it leaves a water surface and enters the air.

Method. speed of light in different media

Light is slower in water and glass than in air (n > 1), so on leaving water for air it speeds up, not down; (c) is not correct. (a), (b) (about 3 x 10^8 m/⁠s) and (d) are correct (C9-A60, C9-A61).

Answer: (c) atmospheric refraction of starlight.

Method. twinkling of stars: atmospheric refraction

Starlight passes through layers of varying refractive index, so the apparent position and brightness of the point-like star flicker (C10-A51, C10-A53).

Answer: (d) Frequency

Method. frequency is fixed by the source, unchanged on refraction

At a boundary speed and wavelength change (v = c/⁠n, lambda' = lambda/⁠n) and the direction changes, but the frequency is the same in both media. Brightness (amplitude) also changes since part is reflected. Standard result (wave refraction); cf. C9-A60.

Answer: (b) go straight.

Method. normal incidence: no bending

At zero angle of incidence sin i = 0, so by Snell's law sin r = 0 and the ray passes undeviated (C9-A57, C9-A58). The ray goes straight, though a part is reflected back.

Answer: (d) both refraction and total internal reflection of light.

Method. mirage: refraction followed by total internal reflection

Mirage arises as rays bend (refraction) through air layers of changing refractive index and then meet a layer at more than the critical angle, undergoing total internal reflection (TP9-A61). Only (d) names both; (c) would omit the refraction.

Answer: (c) refraction of light

Method. refraction at curved glass-water surface magnifies

Light from the lemon bends at the water-glass-air boundaries and the curved tumbler acts like a convex lens, so the lemon looks bigger; this is refraction (C9-A50, C9-A52).

Answer: (a) v_q > v_s

Method. v = c/⁠n; higher index means lower speed

n_q = 1.46 < n_s = 1.77, so v_q = c/⁠1.46 > v_s = c/⁠1.77. The ratio v_q/⁠v_s = 1.77/⁠1.46 = 1.21, not 2, so (d) is wrong (C9-A60, C9-A61).

Answer: (d) successive total internal reflections occur as a ray moves through the fibre.

Method. optical fibre guides light by total internal reflection

The core has higher refractive index than the cladding, and rays strike the core-cladding boundary beyond the critical angle, so they are totally reflected again and again along the fibre (TP9-A28).

Answer: (b) atmospheric refraction

Method. atmospheric refraction raises the apparent position of the Sun

The atmosphere's refractive index increases toward the ground, so light from the Sun below the horizon is bent and the Sun is seen about 2 minutes before actual sunrise and after actual sunset (C10-A49, C10-A55). The apparent shift in position is the effect; the cause is atmospheric refraction.

Answer: (a) v₁ is 1.5 times v₂.

Method. refractive index inversely proportional to speed, n = c/⁠v

n2/⁠n1 = (c/⁠v2)/⁠(c/⁠v1) = v1/⁠v2 = 1.5, so v1 is 1.5 times v2 (C9-A61).

Answer: (d) It is always greater than one

Method. refractive index n = c/⁠v > 1

n = speed in air (vacuum)/⁠speed in the medium; light is slower in every ordinary material, so n > 1 (C9-A61, C9-A62, C9-A101). It cannot be negative, zero or unity for all materials (n = 1 only for vacuum/⁠air itself).

Answer: (d) Total internal reflection

Method. mirage: refraction through hot air layers, then total internal reflection

In desert heat the air next to the ground is hotter and optically rarer, so rays from the sky go down through layers of falling refractive index, meet the next layer beyond the critical angle and are totally internally reflected upward; the eye sees an inverted 'pool' of sky (NCERT TP9-A61, TIR examples include mirage). Of the four options only total internal reflection is the principle; reflection alone, interference and dispersion do not explain it.

Answer: (d) 8 : 9

Method. speed of light in a medium v = c/⁠n

v = c/⁠n, so v_glass/⁠v_water = n_water/⁠n_glass = (4/⁠3)/⁠(3/⁠2) = 8/⁠9, i.e. 8 : 9 (C9-A61, C9-A62).

Answer: (d) 3u = 4R

Method. lens-maker's formula; object and real image at the same distance means u = v = 2f

Convention: Cartesian, first face radius R1 = +R (biconvex), second face R2 = −⁠2R (ratio 1 : 2). 1/⁠f = (n − 1)(1/⁠R1 − 1/⁠R2) = (2 − 1)(1/⁠R + 1/⁠(2R)) = 3/⁠(2R), so f = 2R/⁠3. A real image at the same distance as the object requires u = v = 2f = 4R/⁠3, i.e. 3u = 4R. Verified by calculation. (The other face with radius R would give the same f.)

Answer: (d) −⁠2.0 D

Method. power P = 1/⁠f, concave lens has negative f

f = −⁠0.5 m (Cartesian convention, concave lens). P = 1/⁠(−⁠0.5) = −⁠2.0 D. Verified by calculation.

Answer: (d) [figure (d)]

Method. ray tracing for a thin convex lens

(a) Rays from the focus must emerge parallel, not converge to the other focus. (b) Not all rays from the tip of an object emerge parallel; they converge at the image tip. (c) Rays from a point at 2f must converge to a point at 2f on the other side, not emerge parallel. (d) A ray through the near focus emerges parallel to the axis and a ray through the centre goes undeviated; they meet at the tip of the inverted image. Only (d) is correct. Standard ray-diagram rules.

Answer: (c) +2.0 m

Method. powers of lenses in combination add: P = P1 + P2, f = 1/⁠P

P = 2.5 + (−⁠2.0) = +0.5 D, so f = 1/⁠0.5 = +2.0 m (convex). Verified by calculation.

Answer: (b) +6 dioptre

Method. powers of lenses in contact add: P = P1 + P2, P = 1/⁠f

P1 = 1/⁠0.50 = +2 D and P2 = 1/⁠0.25 = +4 D (convex, f positive). Net P = 2 + 4 = +6 D (f in metres). Standard result for thin lenses in contact.

Answer: (c) 0.5

Method. thin-lens formula 1/⁠v − 1/⁠u = 1/⁠f and m = v/⁠u, Cartesian sign convention

Convention: distances measured from the lens, positive along the incident light. Concave lens: f = −⁠10 cm. Its image is virtual and on the object side, so v = −⁠5 cm. 1/⁠u = 1/⁠v − 1/⁠f = −⁠1/⁠5 + 1/⁠10 = −⁠1/⁠10, so u = −⁠10 cm. m = v/⁠u = (−⁠5)/⁠(−⁠10) = +0.5 (erect, diminished). Verified by calculation.

Answer: (c) +4 Dioptre

Method. power of a lens P = 1/⁠f (f in metres)

f = 25 cm = 0.25 m, so P = 1/⁠0.25 = +4 D (convex lens, f positive). Standard result P = 1/⁠f; one dioptre = 1 m^-1.

Answer: (c) The lens is convex and has a focal length of 0.5 metre

Method. P = 1/⁠f with positive power for a convex lens

f = 1/⁠P = 1/⁠2.0 = 0.5 m; positive power means a convex (converging) lens (C9-A92, C9-A94).

Answer: (b) A concave lens produces both real and virtual images.

Method. image formation by convex and concave lenses

A concave lens always gives a virtual, erect, diminished image (C9-A83), so (b) is the incorrect statement and (d) is correct. A convex lens gives real or virtual images, and smaller, equal or larger according to object position (C9-A75 to C9-A80), so (a) and (c) are correct.

Answer: (a) +2 dioptre

Method. power P = 1/⁠f (metres), convex lens positive

f = 50 cm = 0.5 m, P = 1/⁠0.5 = +2 D (C9-A92, C9-A94). The page prints (b) as +0.02 D, the value from using f in centimetres; (a) is the only correct one.

Answer: (d) 0.25 m

Method. powers of thin lenses in contact add, P = 1/⁠f

P = 2 + 2 = 4 D (C9-A95, TP9-A42), f = 1/⁠P = 1/⁠4 = 0.25 m.

Answer: (a) + 7.5 D

Method. lens-maker's formula, P = (n − 1)(1/⁠R1 − 1/⁠R2), Cartesian convention

Double convex lens: R1 = +10 cm = +0.10 m, R2 = -20 cm = -0.20 m (Cartesian convention, light travelling left to right), n = 1.5 (the page prints '1.5' with a letter l, read as 1.5). P = 0.5 (1/⁠0.10 + 1/⁠0.20) = 0.5 x 15 = +7.5 D (TP9-A32, TP9-A40). Positive, since a convex lens converges.

Answer: (d) [figure (d)]

Method. total internal reflection in an equilateral prism, normal incidence

Normal entry on the left face means no bend; the ray meets the right face at angle of incidence equal to the apex angle, 60 degrees. For glass the critical angle is about 42 degrees (n = 1.5), so there is TIR. With the geometry (axes: apex up) the ray direction is 30 degrees below horizontal and the reflected direction is (0, -1), straight down: perpendicular to the base, so it leaves the base undeviated. (a) and (b) show wrong reflected directions. (c) and (d) both show this vertical path; with the ray entering near the middle of the left face the geometry puts it onto the right face just above the base, as drawn in (d). NCERT TP9-A22, TP9-A27.

Answer: (a) The blue colour is scattered more compared to the red colour

Method. Rayleigh scattering, intensity ∝ 1/⁠λ⁴

Scattering of light by the air molecules varies as 1/⁠λ⁴. Blue (shorter wavelength) is scattered much more than red, which is why the sky is blue. Standard result (Rayleigh scattering).

Answer: (b) 1 and 3

Method. primary rainbow: two refractions and one internal reflection; it is the inner (brighter) bow

In the primary rainbow sunlight is refracted on entering the drop, reflected once inside, and refracted on leaving, so statement 1 is true and 2 is false. The primary bow is the inner, brighter bow (the secondary, with two internal reflections, lies outside it), so statement 3 is true and 4 (more than one reflection) is false. Statements 1 and 3: (b). Standard result.

Answer: (b) dust particles in the air scatter light into our eyes

Method. scattering of light by dust (Tyndall effect)

The beam is visible from the side because dust particles in the air scatter part of the light in all directions, including into our eyes. This is the Tyndall effect (standard result). Reflection from irregular particles is loosely scattering, but 'scatter' is the standard term.

Answer: (d) Blue light deviates the most because blue light has lowest speed in prism

Method. dispersion: n larger for shorter wavelength, v = c/⁠n smaller, deviation larger

Blue light has a shorter wavelength than red, so the glass has a larger refractive index for it, and its speed in the prism v = c/⁠n is the lowest. Deviation δ = (n − 1)A is therefore larger for blue. Red deviates least and travels fastest, so (a) and (c) are wrong; (b) gives the right colour but the wrong reason (blue is slowest, not fastest). Standard result.

Answer: (a) scattering of light by the colloidal particles.

Method. Tyndall effect = scattering by colloidal particles

C10-A59: the scattering of light by colloidal particles, which makes the beam's path visible, is the Tyndall effect.

Answer: (a) [figure (a)]

Method. refraction at the slanted face, then at the vertical face of a right-angled prism

Entering the slanting face (air to glass) the ray meets it obliquely, so it bends towards the normal, that is, downward; (b) and (c) show no bending there and are wrong. At the vertical face the normal is horizontal and the ray now travels slightly downward inside glass, so on emerging (glass to air) it bends away from the normal, further downward. (d) shows no bending at exit and is wrong; (a) shows both bends correctly. Source: Snell's law, n sin i = n' sin r.

Answer: (c) Violet light bends the most, as the refractive index of glass for violet light is greatest.

Method. dispersion: deviation grows with refractive index, and n is greatest for violet

In glass the refractive index is greatest for violet (shortest wavelength) and least for red. Deviation δ = (n − 1)A, so violet bends the most and red the least. Only (c) states both facts correctly (standard result, Cauchy's relation).

Answer: (d) Yellow

Method. additive mixing of coloured light

Red light plus green light gives yellow (additive colour mixing); red + blue gives magenta, not this pair. Standard result.

Answer: (b) Blue, Red and Green

Method. primary colours of light are red, green and blue

Red, green and blue, which combine additively to make other colours including white. Standard result. (Yellow is a secondary colour; violet and indigo are spectral colours.)

Answer: (b) Violet light will deviate the most and it is because of the refraction of light.

Method. dispersion of white light by a prism is due to refraction

Different colours have different refractive indices, so they bend by different amounts on refraction at the two faces; violet bends the most, red the least (C10-A41, C10-A42). Reflection plays no part.

Answer: (b) Raman effect.

Method. Raman effect: inelastic scattering with changed frequency

In Rayleigh scattering the frequency is unchanged; in the Raman effect the scattered light has changed frequency because the molecule gains or loses energy. Standard result (C. V. Raman, 1928).

Answer: (c) Isaac Newton

Method. Newton first obtained the spectrum of sunlight with a prism

C10-A43: Isaac Newton first used a glass prism to obtain the spectrum of sunlight. Raman studied scattering, Rayleigh scattering, Chandrasekhar stellar structure.

Answer: (c) Scattering

Method. scattering: reddening of the Sun

Near the horizon sunlight travels a long path through the atmosphere; blue and short wavelengths are scattered away, leaving the red (C10-A57, C10-A66).

Answer: (a) a shower of rain drops

Method. dispersion by raindrops spreads (disperses) sunlight

Sunlight is spread into its colours (a spectrum) by dispersion in a shower of raindrops, as in a rainbow (C10-A46, C10-A48). A plane mirror only reflects, a convex lens converges, and a convex-plus-concave combination does not split light into colours.

Answer: (b) Red

Method. deviation in a prism depends on wavelength

Red has the longest wavelength and the smallest refractive index in glass, so it deviates the least; violet the most (C10-A42).

Answer: (a) Dispersion of light

Method. rainbow = dispersion by water drops

Raindrops act as small prisms: refraction, internal reflection and dispersion of sunlight form the rainbow (C10-A46, C10-A48).

Answer: (a) larger focal length and larger aperture

Method. telescope design: large objective for light gathering, small eyepiece

In an astronomical telescope the objective has a long focal length and a large aperture to collect more light and give large magnification (m = f_o/⁠f_e); the eyepiece has a short focal length and a small aperture. Standard result.

Answer: (c) Myopia

Method. myopia

A person who can see near objects clearly but not distant ones is short-sighted, i.e. suffers from myopia. Hypermetropia is the reverse; presbyopia affects near vision with age. Standard result.

Answer: (a) The person can see near objects clearly

Method. myopia (short-sightedness)

A myopic eye focuses distant objects in front of the retina; the near point is normal, so near objects are seen clearly and distant ones are blurred (corrected by a concave lens). So (a) is correct and (b) is wrong. Standard result.

Answer: (d) variable focal length and variable aperture size.

Method. the eye lens (accommodation) and the iris/⁠pupil (aperture)

In the eye the ciliary muscles change the curvature of the lens, so its focal length varies (accommodation), and the iris changes the pupil size, so the aperture varies. A camera lens has fixed focal length unless adjusted. Standard result.

Answer: (d) Retina

Method. parts of a compound microscope

A compound microscope has a mirror (illuminator), stage, clips (to hold the slide), objective and eyepiece lenses. The retina is part of the eye, not the microscope (standard reference).

Answer: (a) two convex lenses.

Method. compound microscope: convex objective and convex eyepiece

A compound microscope is a combination of two convex lenses of short focal length, objective and eyepiece. A concave lens or convex mirror is not used in the basic arrangement (standard result).

Answer: (c) [graph (c)]

Method. fixed image distance v on the retina: 1/⁠f = 1/⁠u + 1/⁠v (magnitudes)

The image always forms on the retina, so v is fixed; this rules out (a) and (b) (v against u) and (d) (v against f), since v is a constant. With v fixed, f = uv/⁠(u + v): f is zero at u = 0, rises steeply with u and levels off to f → v for large u. This is graph (c). Derived from the lens formula; the graph is judged from its described shape.

Answer: (b) –⁠0.5 D

Method. myopia corrected by a concave lens whose focal length equals the far point

The far point is 2 m, so the lens must form images of distant objects at 2 m on the same side: f = −⁠2 m (Cartesian convention, concave lens). P = 1/⁠f = −⁠0.5 D. Standard result for myopia correction.

Answer: (c) Newtonian telescope

Method. reflecting (Newtonian) telescope uses mirrors for light collection

The Galilean and Keplerian telescopes are refractors (objective and eyepiece are lenses). The Schmidt telescope uses a spherical mirror plus a glass corrector plate (a lens-like element). The Newtonian telescope has a concave primary mirror and a plane diagonal mirror; its image-forming optics are all mirrors, so it is the telescope meant (standard reference). Strictly the eyepiece, in any telescope, is a lens.

Answer: (d) accommodation of eye

Method. accommodation of the eye

The eye lens changes its focal length (ciliary muscles) so that objects at different distances are focused on the retina; this is accommodation. Far-sightedness and near-sightedness are defects, not the mechanism (standard result; the pupil's variable aperture similarly handles illumination).

Answer: (c) distant object is focused before the retina.

Method. myopia: image of distant objects falls in front of the retina

In myopia the eye lens is too curved or the eyeball too long, so a distant object's image forms in front of the retina (C10-A22, C10-A23); nearby objects are seen clearly (C10-A20).

Answer: (c) 2 cm

Method. telescope magnification m = f_o/⁠f_e

f_e = f_o/⁠m = 50/⁠25 = 2 cm (TP9-A55, TP9-A56).

Answer: (d) 1, 2 and 3

Method. microscope and telescope: focal lengths and magnification

(1) True: both are convex lenses (TP9-A52). (2) True: telescope objective has a large focal length, eyepiece small (TP9-A55). (3) True: m = f_o/⁠f_e, increases with f_o (TP9-A56). (4) False: microscope m ~ L/⁠f_o, falls as f_o increases (TP9-A53). So 1, 2 and 3.

Answer: (a) Human eye is a refracting system containing a diverging lens.

Method. structure of the human eye

The eye lens is a converging (convex) lens, not a diverging one; the cornea and lens refract light onto the retina, so (a) is not correct (C10-A02, C10-A07). The retina holds light-sensitive cells that send electrical signals (C10-A08, C10-A09); the image on the retina is real and inverted (C10-A07); two eyes give depth judgement (standard fact). (b), (c), (d) are correct.

Answer: (a) 4 2 1 3

Method. remedy for each defect of vision

Hypermetropia is corrected by a convex lens (C10-A29) -> 4. Presbyopia needs bi-focal lenses (C10-A30, C10-A31) -> 2. Myopia is corrected by a concave lens (C10-A24) -> 1. Cataract is treated by surgery (C10-A19) -> 3. Code A-4, B-2, C-1, D-3.

Answer: (d) Magnification decreases with the increase in focal length of eyepiece.

Method. compound microscope magnification M ~ (L/⁠f_o)(D/⁠f_e)

For a compound microscope m = (L/⁠f_o)(1 + D/⁠f_e) (TP9-A53); with f_o and tube length fixed, m varies inversely with f_e, so magnification decreases as the eyepiece focal length increases. Option (b) is also wrong: m falls as f_o grows.