IOE entrance Physics · Chapter 3
Geometric and Physical Optics
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56 questions in 7 syllabus topics · 23 are 2-mark questions.
3.1 Reflection
8 questions
1. A concave mirror has a radius of curvature of 40 cm. Its focal length is:
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Relate focal length to radius of curvature.
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Answer: C. 20 cm
For a spherical mirror f = R/2, so f = 40/2 = 20 cm.
2. For a real object placed anywhere in front of a convex mirror, the image is always:
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Think of how a convex mirror spreads reflected rays.
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Answer: D. virtual, erect and diminished
A convex mirror diverges light, so the reflected rays only appear to meet behind it: the image is virtual, erect and smaller than the object, lying between the pole and the focus.
3. An object is placed 30 cm in front of a concave mirror of focal length 20 cm. The image formed is:
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Use the mirror formula, then m = −v/u.
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Answer: D. 60 cm in front of the mirror, real, inverted and twice the size
Using 1/v + 1/u = 1/f with real distances positive: 1/v = 1/20 − 1/30 = 1/60, so v = 60 cm (real, in front). Magnification m = −v/u = −60/30 = −2, so it is inverted and twice the size.
4. A ray of light falls on a plane mirror. Keeping the incident ray fixed, the mirror is rotated through an angle θ. The reflected ray turns through:
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Both the angle of incidence and the angle of reflection change.
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Answer: B. 2θ
Rotating the mirror by θ changes the angle of incidence by θ, so the angle of reflection also changes by θ; the reflected ray turns by θ + θ = 2θ.
5. Two plane mirrors are inclined to each other at 60°. The number of images of a point object placed symmetrically between them is:
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Use n = 360°/θ − 1 when 360/θ is even.
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Answer: A. 5
n = 360°/θ − 1 = 360/60 − 1 = 5 (since 360/θ is even).
6. An object is placed 15 cm in front of a convex mirror of focal length 15 cm. The magnification produced is:
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Apply 1/v + 1/u = 1/f with proper signs, then m = −v/u.
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Answer: D. +0.5
Sign convention: u = −15 cm, f = +15 cm. 1/v = 1/f − 1/u = 1/15 + 1/15 = 2/15, so v = +7.5 cm (behind the mirror). m = −v/u = −7.5/(−15) = +0.5, an erect, half-size virtual image.
7. A concave mirror forms a real image, three times the size of the object, on a screen kept 60 cm from the mirror. The focal length of the mirror is:
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Find u from the magnification first.
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Answer: A. 15 cm
Image distance v = 60 cm and |m| = v/u = 3, so u = 20 cm. Then 1/f = 1/60 + 1/20 = 4/60, giving f = 15 cm.
8. Convex mirrors are preferred as rear-view mirrors in vehicles mainly because they:
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Think about how much of the road behind the driver can see.
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Answer: C. give an erect image with a wider field of view
A convex mirror always gives an erect, diminished image and covers a much wider region behind the vehicle than a plane mirror of the same size.
3.2 Refraction, prisms and lenses
10 questions
9. The refractive index of glass is 1.5. The speed of light in this glass is:
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Refractive index is the ratio c/v.
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Answer: A. 2 × 10⁸ m/s
v = c/μ = (3 × 10⁸)/1.5 = 2 × 10⁸ m/s.
10. The critical angle for light going from a medium of refractive index √2 into air is:
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Use sin C = 1/μ.
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Answer: D. 45°
sin C = 1/μ = 1/√2, so C = 45°.
11. A coin lies at the bottom of a beaker filled with water to a depth of 20 cm (μ of water = 4/3). When viewed normally from above, the coin appears to be raised by:
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Apparent depth = real depth ÷ μ.
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Answer: D. 5 cm
Apparent depth = real depth/μ = 20 × 3/4 = 15 cm, so the apparent rise = 20 − 15 = 5 cm.
12. A prism of angle 60° gives a minimum deviation of 30°. The refractive index of its material is:
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Use the prism formula at minimum deviation.
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Answer: D. √2
μ = sin((A + δm)/2)/sin(A/2) = sin 45°/sin 30° = (1/√2)/(1/2) = √2.
13. An object is placed 20 cm in front of a concave lens of focal length 20 cm. The image is formed:
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Use the lens formula with f negative for a concave lens.
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Answer: B. 10 cm from the lens on the same side as the object
Using 1/v − 1/u = 1/f with u = −20 cm and f = −20 cm: 1/v = −1/20 − 1/20 = −1/10, so v = −10 cm. The negative sign means the virtual image is on the same side as the object.
14. An equiconvex glass lens (μ = 1.5) has both radii of curvature equal to 20 cm. When it is completely immersed in water (μ = 4/3), its focal length becomes:
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Use the lens maker's formula with the relative refractive index of glass with respect to water.
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Answer: C. 80 cm
In water, relative index = 1.5/(4/3) = 9/8. 1/f = (9/8 − 1)(1/20 + 1/20) = (1/8)(1/10) = 1/80, so f = 80 cm (it was 20 cm in air).
15. A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 30 cm. The combination behaves as a:
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Add the powers (1/f) with a sign for each lens.
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Answer: C. converging lens of focal length 60 cm
1/F = 1/f₁ + 1/f₂ = 1/20 + 1/(−30) = (3 − 2)/60 = 1/60, so F = +60 cm, a converging combination.
16. The working of an optical fibre is based on:
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What keeps light trapped inside the core?
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Answer: B. total internal reflection
Light entering the core strikes the core–cladding boundary at angles greater than the critical angle and is totally internally reflected again and again along the fibre.
17. Total internal reflection can take place only when light travels from:
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Two conditions: direction of travel and angle size.
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Answer: B. a denser to a rarer medium with angle of incidence greater than the critical angle
Only when going from optically denser to rarer can the refracted ray bend away so much that, beyond the critical angle, no refraction is possible and all light is reflected.
18. A ray of light is incident at 60° on one face of a glass slab of thickness 6 cm and refractive index √3. The lateral shift of the emergent ray is:
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Find r by Snell's law, then use d = t sin(i − r)/cos r.
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Answer: B. 2√3 cm
sin r = sin 60°/√3 = 1/2, so r = 30°. Lateral shift d = t sin(i − r)/cos r = 6 × sin 30°/cos 30° = 6 × (1/2)/(√3/2) = 6/√3 = 2√3 cm.
3.3 Dispersion and scattering
8 questions
19. If μv, μr and μy are the refractive indices of a material for violet, red and yellow (mean) light, the dispersive power of the material is:
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It is angular dispersion divided by the mean deviation.
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Answer: A. (μv − μr)/(μy − 1)
Dispersive power ω = angular dispersion/mean deviation = (μv − μr)A/((μy − 1)A) = (μv − μr)/(μy − 1).
20. For a glass, μv = 1.66 and μr = 1.62, and the mean refractive index is the average of these two. The dispersive power of the glass is:
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Divide (μv − μr) by (μy − 1), not by μy.
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Answer: A. 0.0625
μy = (1.66 + 1.62)/2 = 1.64. ω = (μv − μr)/(μy − 1) = 0.04/0.64 = 0.0625.
21. When white light passes through a glass prism, the colour that is deviated the least is:
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Which colour has the smallest refractive index in glass?
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Answer: C. red
Refractive index of glass is smallest for the longest wavelength (red), so red is deviated least and violet the most.
22. The blue colour of the clear sky is mainly due to:
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Think of Rayleigh's law.
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Answer: B. scattering of light, which varies as 1/λ⁴
By Rayleigh's law scattered intensity ∝ 1/λ⁴, so short-wavelength blue light is scattered far more than red and reaches the eye from all over the sky.
23. According to Rayleigh's law, the ratio of the intensity of scattered light of wavelength 400 nm to that of wavelength 600 nm (same incident intensity) is:
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Scattered intensity varies as the inverse fourth power of wavelength.
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Answer: D. 81/16
I ∝ 1/λ⁴, so I₄₀₀/I₆₀₀ = (600/400)⁴ = (3/2)⁴ = 81/16 ≈ 5.
24. An achromatic convergent doublet of focal length 40 cm is to be made from a crown glass lens (ω = 0.02) and a flint glass lens (ω = 0.04) in contact. The focal lengths of the crown and flint lenses are respectively:
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Combine ω₁/f₁ + ω₂/f₂ = 0 with 1/F = 1/f₁ + 1/f₂.
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Answer: C. +20 cm and −40 cm
Achromatism: ω₁/f₁ + ω₂/f₂ = 0 gives f₂ = −2f₁. Then 1/F = 1/f₁ − 1/(2f₁) = 1/(2f₁) = 1/40, so f₁ = +20 cm and f₂ = −40 cm.
25. Chromatic aberration in a lens arises because:
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Which property of the glass depends on colour?
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Answer: C. the refractive index, and hence focal length, of the lens differs for different colours
Since μ depends on wavelength, f also depends on wavelength (violet focuses nearer than red), so white light does not come to a single focus.
26. Spherical aberration of a convex lens can be reduced by:
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Which rays are responsible for spherical aberration?
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Answer: A. using a stop that cuts off the marginal rays
Spherical aberration occurs because marginal and paraxial rays focus at different points. Blocking the marginal rays with a stop reduces it (at the cost of brightness). The achromatism condition and monochromatic light address chromatic, not spherical, aberration.
3.4 Nature and propagation of light
6 questions
27. According to Huygens' principle:
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Recall how a new wavefront is constructed.
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Answer: D. every point on a wavefront acts as a source of secondary wavelets
Huygens proposed that each point on a wavefront emits secondary wavelets travelling at the wave speed; the forward envelope (common tangent) of these wavelets gives the new wavefront.
28. The wavefront of light reaching the earth from a distant star is:
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Consider a tiny patch of a sphere of enormous radius.
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Answer: D. plane
A point source produces spherical wavefronts, but at very large distances a small part of a huge sphere is effectively flat, i.e. a plane wavefront.
29. According to Newton's corpuscular theory, the speed of light in water compared with that in air is:
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How did the corpuscular theory explain bending towards the normal?
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Answer: B. greater
The corpuscular theory explained refraction towards the normal by an attractive force increasing the normal component of velocity, so it predicted a higher speed in water. Foucault's measurement showed it is actually smaller, supporting the wave theory.
30. In Michelson's rotating-mirror method an octagonal mirror is used, and the distance between the rotating mirror and the distant reflecting mirror is 37.5 km. A steady image is first obtained when the octagonal mirror rotates at 500 rev/s. The speed of light obtained is:
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Light covers 2D while the octagon turns one-eighth of a revolution.
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Answer: D. 3 × 10⁸ m/s
A steady image needs the mirror to turn 1/8 rev while light travels 2D. Time = 1/(8n) = 2D/c, so c = 16nD = 16 × 500 × 37 500 = 3 × 10⁸ m/s.
31. The time taken by light to cross normally a glass slab of thickness 3 cm and refractive index 1.5 is:
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Find the speed in glass first.
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Answer: B. 1.5 × 10⁻¹⁰ s
Speed in glass v = c/μ = 2 × 10⁸ m/s. t = d/v = 0.03/(2 × 10⁸) = 1.5 × 10⁻¹⁰ s.
32. When light passes from air into glass, the quantity that remains unchanged is its:
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Which property is fixed by the source?
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Answer: B. frequency
Frequency is set by the source and does not change on refraction; speed and wavelength both decrease by the factor μ.
3.5 Interference
8 questions
33. Two sources of light are said to be coherent if they have:
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What must stay constant for a stable fringe pattern?
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Answer: C. the same frequency and a constant phase difference
Sustained interference needs sources with the same frequency (wavelength) whose phase difference stays constant in time.
34. In Young's double slit experiment with slit separation d, screen distance D and wavelength λ, the fringe width is:
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Fringe width increases with D and decreases with d.
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Answer: A. λD/d
Successive bright fringes are separated by β = λD/d.
35. In Young's double slit experiment, the slits are 0.3 mm apart and the screen is 1.5 m away. If light of wavelength 600 nm is used, the fringe width is:
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Convert all lengths to metres and use β = λD/d.
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Answer: B. 3 mm
β = λD/d = (600 × 10⁻⁹ × 1.5)/(0.3 × 10⁻³) = 3 × 10⁻³ m = 3 mm.
36. In a Young's double slit experiment in air the fringe width is 4 mm. If the whole apparatus is immersed in water (μ = 4/3), the fringe width becomes:
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What happens to the wavelength in water?
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Answer: A. 3 mm
In water λ' = λ/μ, so β' = β/μ = 4 × 3/4 = 3 mm.
37. The intensities of light from two coherent slits are in the ratio 9 : 1. The ratio of maximum to minimum intensity in the interference pattern is:
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Convert intensities to amplitudes first.
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Answer: C. 4 : 1
Amplitude ratio = √9 : √1 = 3 : 1. Imax/Imin = (3 + 1)²/(3 − 1)² = 16/4 = 4.
38. In a double slit experiment, d = 0.5 mm, D = 1 m and λ = 500 nm. The distance of the third dark fringe from the central bright fringe is:
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Dark fringes lie halfway between bright ones.
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Answer: B. 2.5 mm
β = λD/d = (500 × 10⁻⁹ × 1)/(0.5 × 10⁻³) = 1 mm. The nth dark fringe is at (2n − 1)β/2, so for n = 3 it is 5 × 1/2 = 2.5 mm.
39. Two separate sodium lamps placed side by side do not produce an observable interference pattern because:
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Think about coherence.
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Answer: B. their light waves do not have a constant phase difference
Light from independent sources is emitted by random atomic transitions, so the phase difference changes billions of times per second and the fringes wash out. The sources are not coherent.
40. If white light is used in Young's double slit experiment, the central fringe is:
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What is the path difference at the centre for each colour?
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Answer: A. white, with coloured fringes on either side
At the centre the path difference is zero for every wavelength, so all colours interfere constructively and the central fringe is white; away from it the fringes of different colours separate.
3.6 Diffraction
8 questions
41. In Fraunhofer diffraction at a single slit of width a, the condition for the nth minimum (n = 1, 2, 3, …) is:
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The first minimum occurs when the path difference across the whole slit is one wavelength.
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Answer: C. a sin θ = nλ
Dividing the slit into 2n equal zones, the contributions cancel in pairs when the path difference across the slit is nλ, giving a sin θ = nλ.
42. Light of wavelength 500 nm falls normally on a slit of width 0.2 mm. The width of the central maximum on a screen 2 m away is:
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The central maximum extends to the first minimum on each side.
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Answer: A. 10 mm
Width of central maximum = 2λD/a = 2 × 500 × 10⁻⁹ × 2/(0.2 × 10⁻³) = 10⁻² m = 10 mm.
43. A diffraction grating has 5000 lines per cm. For light of wavelength 500 nm at normal incidence, the highest order spectrum that can actually be observed is:
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Use d sin θ = nλ with sin θ < 1.
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Answer: B. 3
d = 1/5000 cm = 2 × 10⁻⁶ m. n = d sin θ/λ ≤ d/λ = 4. For n = 4, sin θ = 1 (θ = 90°), which lies along the grating and cannot be observed, so the highest observable order is 3.
44. Light of wavelength 500 nm falling normally on a grating gives its first order maximum at 30°. The number of lines per centimetre on the grating is:
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Find the grating element d from d sin θ = nλ.
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Answer: C. 10 000
d sin θ = λ gives d = 500 × 10⁻⁹/0.5 = 10⁻⁶ m = 10⁻⁴ cm. Lines per cm = 1/d = 10 000.
45. The resolving power of a telescope can be increased by:
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Resolving power ∝ D/λ.
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Answer: A. increasing the diameter of the objective
Resolving power of a telescope = D/(1.22λ), so a larger objective aperture (or shorter wavelength) resolves finer detail. The eyepiece affects magnification, not resolution.
46. In single slit diffraction, if the slit width is decreased, the central maximum:
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Angular width is inversely proportional to slit width.
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Answer: C. becomes wider
The angular half-width of the central maximum is θ ≈ λ/a, so a smaller slit width a gives a wider central maximum.
47. According to Rayleigh's criterion, two point sources are just resolved when:
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Think of the limit just before two patterns merge.
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Answer: C. the central maximum of one image falls on the first minimum of the other
Rayleigh's criterion: the images are just resolved when the principal maximum of one diffraction pattern coincides with the first minimum of the other.
48. The minimum number of lines a grating must have to just resolve the sodium lines of wavelengths 589.0 nm and 589.6 nm in the first order is about:
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Use λ/Δλ = nN.
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Answer: B. 982
Resolving power λ/Δλ = nN. With n = 1: N = 589/0.6 ≈ 982.
3.7 Polarization
8 questions
49. The phenomenon of polarization shows that light waves are:
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Which kind of wave has vibrations that can be confined to one plane?
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Answer: C. transverse
Only transverse waves have vibrations perpendicular to the direction of travel that can be restricted to a single plane, so polarization proves light is transverse.
50. Brewster's law relates the polarizing angle θp to the refractive index μ as:
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It involves the tangent of the angle.
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Answer: A. μ = tan θp
At the polarizing angle the reflected light is completely plane polarized, and Brewster found μ = tan θp.
51. Unpolarized light is incident on a glass surface (μ = √3) at the polarizing angle. The angle of refraction is:
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Find θp, then use θp + r = 90°.
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Answer: D. 30°
tan θp = √3 gives θp = 60°. At Brewster's angle the reflected and refracted rays are perpendicular, so r = 90° − 60° = 30°.
52. Unpolarized light of intensity I₀ passes through a polarizer and then through an analyzer whose axis makes 30° with that of the polarizer. The intensity of the emergent light is:
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First halve the intensity, then apply Malus' law.
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Answer: A. 3I₀/8
The polarizer transmits I₀/2. By Malus' law the analyzer transmits (I₀/2)cos²30° = (I₀/2)(3/4) = 3I₀/8.
53. Plane polarized light of intensity I falls on an analyzer. For the transmitted intensity to be I/4, the angle between the plane of polarization and the analyzer axis must be:
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Use I = I₀cos²θ.
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Answer: B. 60°
Malus' law: I cos²θ = I/4, so cos θ = 1/2 and θ = 60°.
54. Sound waves in air cannot be polarized because they are:
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Polarization needs vibrations perpendicular to propagation.
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Answer: B. longitudinal
In a longitudinal wave the vibrations are along the direction of propagation, so there is no transverse direction of vibration to select; such waves cannot be polarized.
55. When light is incident on a transparent surface at Brewster's angle, the angle between the reflected ray and the refracted ray is:
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Combine Brewster's law with Snell's law.
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Answer: C. 90°
tan θp = μ = sin θp/sin r gives sin r = cos θp, so θp + r = 90°, making the reflected and refracted rays mutually perpendicular.
56. Two polaroids are crossed so that no light passes. A third polaroid is inserted between them with its axis at 45° to each. If unpolarized light of intensity I₀ falls on the first polaroid, the intensity finally emerging is:
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Apply Malus' law at each of the last two polaroids.
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Answer: A. I₀/8
After the first: I₀/2. After the middle one (45°): (I₀/2)cos²45° = I₀/4. After the last (45°): (I₀/4)cos²45° = I₀/8.