NCERT Class X Physics

Chapter 2: Human Eye and Colourful World

100 Multiple Choice Questions Workbook

Topics Covered: Structure of Human Eye, Defects of Vision, Atmospheric Refraction, Scattering of Light

SECTION A: QUESTIONS
SIMPLEST LEVEL (Questions 1-15)
Basic recall and recognition
1. The human eye forms image on:
(a) Cornea     (b) Iris     (c) Retina     (d) Pupil
2. The coloured part of the eye is called:
(a) Cornea     (b) Iris     (c) Pupil     (d) Lens
3. The hole in the iris is called:
(a) Cornea     (b) Retina     (c) Pupil     (d) Lens
4. The transparent front part of the eye is:
(a) Iris     (b) Cornea     (c) Pupil     (d) Retina
5. Myopia is also known as:
(a) Far-sightedness     (b) Near-sightedness     (c) Old age vision     (d) Color blindness
6. Hypermetropia is corrected using:
(a) Concave lens     (b) Convex lens     (c) Bifocal lens     (d) Plane mirror
7. The sky appears blue due to:
(a) Reflection     (b) Refraction     (c) Scattering     (d) Absorption
8. The sun appears red at sunset due to:
(a) Refraction     (b) Scattering     (c) Reflection     (d) Dispersion
9. The minimum distance of distinct vision for a normal eye is:
(a) 15 cm     (b) 20 cm     (c) 25 cm     (d) 30 cm
10. Presbyopia occurs due to:
(a) Old age     (b) Birth defect     (c) Accident     (d) Disease
11. Which part of eye controls the amount of light entering?
(a) Cornea     (b) Lens     (c) Iris     (d) Retina
12. Twinkling of stars is due to:
(a) Atmospheric refraction     (b) Scattering     (c) Reflection     (d) Absorption
13. Rainbow is formed due to:
(a) Refraction only     (b) Dispersion only     (c) Refraction and dispersion     (d) Scattering
14. The image formed on retina is:
(a) Real and inverted     (b) Virtual and erect     (c) Real and erect     (d) Virtual and inverted
15. Astigmatism is corrected using:
(a) Concave lens     (b) Convex lens     (c) Cylindrical lens     (d) Bifocal lens
SIMPLE LEVEL (Questions 16-35)
Basic understanding and application
16. The change in focal length of eye lens is called:
(a) Accommodation     (b) Adaptation     (c) Adjustment     (d) Aberration
17. A person cannot see objects beyond 2 m clearly. He suffers from:
(a) Hypermetropia     (b) Myopia     (c) Presbyopia     (d) Astigmatism
18. The power of accommodation decreases with:
(a) Age     (b) Height     (c) Weight     (d) Exercise
19. Which color has the maximum wavelength in visible spectrum?
(a) Violet     (b) Blue     (c) Green     (d) Red
20. Scattering of light is maximum for:
(a) Red light     (b) Green light     (c) Blue light     (d) Yellow light
21. The far point of a myopic eye is:
(a) At infinity     (b) At 25 cm     (c) Less than infinity     (d) At focus
22. The near point of hypermetropic eye is:
(a) At 25 cm     (b) Less than 25 cm     (c) More than 25 cm     (d) At infinity
23. Advanced sunrise and delayed sunset are due to:
(a) Scattering     (b) Atmospheric refraction     (c) Reflection     (d) Dispersion
24. The danger signals are red because:
(a) Red is scattered most     (b) Red is scattered least     (c) Red is bright     (d) Red is attractive
25. Bifocal lens is used to correct:
(a) Myopia only     (b) Hypermetropia only     (c) Presbyopia     (d) Astigmatism
26. The ciliary muscles help in:
(a) Changing pupil size     (b) Changing lens curvature     (c) Moving the eye     (d) Blinking
27. Persistence of vision is about:
(a) 1/10 second     (b) 1/16 second     (c) 1/20 second     (d) 1/25 second
28. The splitting of white light into component colors is called:
(a) Reflection     (b) Refraction     (c) Dispersion     (d) Scattering
29. Rayleigh scattering is inversely proportional to:
(a) λ     (b) λ²     (c) λ³     (d) λ⁴
30. The least distance of distinct vision is also called:
(a) Far point     (b) Near point     (c) Focus     (d) Range
31. Atmospheric refraction occurs due to:
(a) Uniform air density     (b) Varying air density     (c) Air pollution     (d) Air pressure
32. The eye defect in which distant objects appear blurred is:
(a) Myopia     (b) Hypermetropia     (c) Presbyopia     (d) Astigmatism
33. Which part of eye is equivalent to diaphragm of camera?
(a) Lens     (b) Cornea     (c) Iris     (d) Retina
34. The refractive index of atmosphere is:
(a) Constant     (b) Variable with height     (c) Zero     (d) Infinite
35. Primary colors are:
(a) Red, Blue, Yellow     (b) Red, Green, Blue     (c) Red, Green, Yellow     (d) Blue, Green, Yellow
EASY LEVEL (Questions 36-55)
Conceptual understanding and simple analysis
36. A person with myopia cannot see clearly beyond 80 cm. The power of lens required is:
(a) -1.25 D     (b) +1.25 D     (c) -0.8 D     (d) +0.8 D
37. The angular size of sun and moon as seen from earth is approximately:
(a) 0.5°     (b) 1°     (c) 1.5°     (d) 2°
38. If the refractive index of water is 4/3, the critical angle is:
(a) 41.8°     (b) 48.6°     (c) 53.1°     (d) 60°
39. Hypermetropia can be corrected by a lens of power:
(a) Negative     (b) Positive     (c) Zero     (d) Infinite
40. The condition in which eye cannot distinguish between certain colors is:
(a) Myopia     (b) Hypermetropia     (c) Color blindness     (d) Presbyopia
41. The phenomenon of total internal reflection is observed in:
(a) Mirage     (b) Rainbow     (c) Blue sky     (d) Red sunset
42. The wavelength of red light in air is 700 nm. Its wavelength in water (μ = 4/3) is:
(a) 525 nm     (b) 700 nm     (c) 933 nm     (d) 280 nm
43. A concave lens of focal length 20 cm is used to correct myopia. The far point is:
(a) 15 cm     (b) 20 cm     (c) 25 cm     (d) 30 cm
44. Stars appear to twinkle but planets do not because:
(a) Stars are farther     (b) Planets are self-luminous     (c) Stars are point sources     (d) Planets are brighter
45. The sun is visible about 2 minutes before actual sunrise due to:
(a) Scattering     (b) Refraction     (c) Reflection     (d) Diffraction
46. At noon, sun appears white because:
(a) Less scattering     (b) More scattering     (c) No atmosphere     (d) Direct path
47. The power of normal eye lens varies from:
(a) 40 D to 44 D     (b) 44 D to 60 D     (c) 60 D to 64 D     (d) 64 D to 68 D
48. The term VIBGYOR represents:
(a) Colors of rainbow     (b) Parts of eye     (c) Types of lenses     (d) Vision defects
49. Cataract can be corrected by:
(a) Lens     (b) Surgery     (c) Exercise     (d) Medicine
50. The speed of light in vacuum is approximately:
(a) 3 × 10⁶ m/s     (b) 3 × 10⁷ m/s     (c) 3 × 10⁸ m/s     (d) 3 × 10⁹ m/s
51. Myopia occurs when:
(a) Eyeball is too short     (b) Eyeball is too long     (c) Lens is too weak     (d) Retina is damaged
52. The intensity of scattered light is proportional to:
(a) 1/λ²     (b) 1/λ³     (c) 1/λ⁴     (d) 1/λ
53. A convex lens of power +2 D has focal length:
(a) 25 cm     (b) 50 cm     (c) 100 cm     (d) 200 cm
54. The phenomenon due to which sun appears flattened at horizon:
(a) Refraction     (b) Scattering     (c) Reflection     (d) Dispersion
55. In normal eye, the image distance from lens is approximately:
(a) 1.7 cm     (b) 2.3 cm     (c) 2.5 cm     (d) 3.0 cm
MODERATE LEVEL (Questions 56-75)
Application and problem-solving
56. A person can read clearly up to 25 cm but cannot see beyond 80 cm. The powers of lenses required for reading and distance vision are:
(a) +4 D, -1.25 D     (b) -4 D, +1.25 D     (c) 0 D, -1.25 D     (d) +1.25 D, 0 D
57. A myopic person has far point at 200 cm. The power of corrective lens is:
(a) -0.5 D     (b) +0.5 D     (c) -2 D     (d) +2 D
58. Critical angle for total internal reflection depends on:
(a) Angle of incidence     (b) Wavelength of light     (c) Refractive indices     (d) Intensity of light
59. A hypermetropic person has near point at 40 cm. Power of correcting lens:
(a) +1.5 D     (b) -1.5 D     (c) +2.5 D     (d) -2.5 D
60. The ratio of intensities of scattered blue and red light is approximately:
(a) 2:1     (b) 4:1     (c) 5:1     (d) 16:1
61. If earth had no atmosphere, the sky would appear:
(a) Blue     (b) Red     (c) White     (d) Black
62. The curvature of cornea in normal eye provides power of about:
(a) 40 D     (b) 44 D     (c) 60 D     (d) 20 D
63. For a person with presbyopia having near point 40 cm and far point 200 cm, the powers needed are:
(a) +1.5 D, -0.5 D     (b) -1.5 D, +0.5 D     (c) +2.5 D, -0.5 D     (d) -2.5 D, +0.5 D
64. The angular diameter of sun is 32 minutes. Its linear diameter is approximately:
(a) 1.39 × 10⁹ m     (b) 1.39 × 10⁸ m     (c) 1.39 × 10¹⁰ m     (d) 1.39 × 10⁷ m
65. Atmospheric refraction causes:
(a) Stars to appear higher     (b) Sun to appear lower     (c) Moon to appear smaller     (d) Planets to twinkle
66. The wavelength range of visible light is approximately:
(a) 400-700 nm     (b) 300-800 nm     (c) 200-900 nm     (d) 100-1000 nm
67. If the power of accommodation is 4 D, the near point is:
(a) 20 cm     (b) 25 cm     (c) 30 cm     (d) 40 cm
68. Mirage is formed due to:
(a) Refraction only     (b) Total internal reflection     (c) Scattering     (d) Dispersion
69. The separation between objective lens and eyepiece in normal adjustment of telescope is:
(a) f₀ + fₑ     (b) f₀ - fₑ     (c) f₀ × fₑ     (d) f₀/fₑ
70. A person uses spectacles of power -2 D. His far point without glasses is:
(a) 25 cm     (b) 50 cm     (c) 100 cm     (d) 200 cm
71. The refractive index of eye lens varies from:
(a) 1.36 to 1.40     (b) 1.40 to 1.44     (c) 1.44 to 1.50     (d) 1.33 to 1.36
72. The condition for total internal reflection is:
(a) i > C, denser to rarer     (b) i < C, rarer to denser     (c) i = C, any medium     (d) i > C, rarer to denser
73. At what time does sun appear most red?
(a) Morning     (b) Noon     (c) Evening     (d) Both morning and evening
74. The magnifying power of simple microscope is maximum when image is formed at:
(a) Infinity     (b) Near point     (c) Far point     (d) Focus
75. Which phenomenon explains the reddish appearance of sun during sunrise and sunset?
(a) Dispersion     (b) Interference     (c) Rayleigh scattering     (d) Diffraction
JUST TOUGH LEVEL (Questions 76-90)
Advanced problem-solving and analysis
76. A person cannot see objects distinctly closer than 40 cm and farther than 200 cm. The powers of bifocal lens are:
(a) +1.5 D and -0.5 D     (b) -1.5 D and +0.5 D     (c) +2.5 D and -0.5 D     (d) +1.5 D and -1 D
77. The dispersive power of crown glass is 0.02. If μᵥ = 1.52 and μᵣ = 1.51, then μᵧ is:
(a) 1.512     (b) 1.515     (c) 1.518     (d) 1.520
78. For normal eye, the distance between lens and retina is about 2.5 cm. If the eye is filled with water (μ = 4/3), the image will form:
(a) On retina     (b) Behind retina     (c) In front of retina     (d) Cannot determine
79. A myopic person has range of vision from 12 cm to 50 cm. Power of lens to see up to infinity:
(a) -2 D     (b) +2 D     (c) -8.33 D     (d) +8.33 D
80. The critical angle for light going from glass (μ = 1.5) to air is approximately:
(a) 30°     (b) 42°     (c) 48.6°     (d) 60°
81. If blue light has wavelength 450 nm in air, its wavelength in water (μ = 4/3) is:
(a) 337.5 nm     (b) 450 nm     (c) 600 nm     (d) 300 nm
82. A compound microscope has objective focal length 1 cm and eyepiece focal length 5 cm. If the distance between them is 30 cm, magnification is:
(a) 120     (b) 150     (c) 144     (d) 100
83. The least count of human eye (resolving power) is about:
(a) 1 arc minute     (b) 2 arc minutes     (c) 30 arc seconds     (d) 1 arc second
84. Rainbow is formed when sun is behind the observer at an angle of:
(a) 40°     (b) 42°     (c) 45°     (d) 50°
85. The power of accommodation for normal eye decreases from 4 D at age 10 to 1 D at age 60. The near point at age 60 is:
(a) 40 cm     (b) 50 cm     (c) 100 cm     (d) 25 cm
86. A person can see clearly from 20 cm to 2 m. The range of power accommodation is:
(a) 4.5 D     (b) 5 D     (c) 4 D     (d) 3.5 D
87. The intensity ratio of light scattered by particles of size comparable to wavelength (Mie scattering) varies as:
(a) λ⁻⁴     (b) λ⁻²     (c) λ⁰     (d) λ²
88. For a person with myopia having far point at 1 m, using a lens of power -1 D, the new far point becomes:
(a) 2 m     (b) Infinity     (c) 50 cm     (d) 1.5 m
89. The ratio of diameter of pupil in bright light to that in dim light is approximately:
(a) 1:2     (b) 1:4     (c) 1:6     (d) 1:8
90. If the earth's atmosphere were twice as dense, the sky would appear:
(a) More blue     (b) Less blue     (c) Same blue     (d) Red
EXAM LEVEL (Questions 91-100)
Comprehensive and challenging
91. A compound microscope has an objective of focal length 2 cm and eyepiece of focal length 5 cm. The object is placed 2.5 cm from objective. The magnifying power when image is formed at 25 cm from eyepiece is:
(a) 50     (b) 60     (c) 75     (d) 100
92. A hypermetropic person has power of accommodation 2 D and near point 50 cm. After using spectacles, his near point becomes 25 cm. The power of spectacles is:
(a) +1 D     (b) +2 D     (c) +1.5 D     (d) +0.5 D
93. The sun subtends an angle of 0.5° at earth. If the distance is 1.5 × 10¹¹ m, the diameter of sun is approximately:
(a) 1.31 × 10⁹ m     (b) 1.31 × 10⁸ m     (c) 1.31 × 10¹⁰ m     (d) 1.31 × 10⁷ m
94. A person uses a lens of power +3 D for reading. His near point without lens is 75 cm. His age-related loss in accommodation is:
(a) 1.67 D     (b) 2.67 D     (c) 3.67 D     (d) 4.67 D
95. A ray of light passes from air to glass (μ = 1.5) at an angle of 30°. The lateral displacement for a glass slab of thickness 6 cm is:
(a) 1.76 cm     (b) 2.24 cm     (c) 2.76 cm     (d) 3.24 cm
96. The power of a thick lens in air is 10 D. When immersed in water (μ = 4/3), its power becomes approximately:
(a) 2.5 D     (b) 3.3 D     (c) 7.5 D     (d) 13.3 D
97. The minimum thickness of soap film (μ = 1.33) for constructive interference of light λ = 600 nm at normal incidence is:
(a) 113 nm     (b) 225 nm     (c) 300 nm     (d) 450 nm
98. A person has presbyopia with near point 100 cm and far point 40 cm. The powers of progressive lens required are:
(a) +3 D to +1 D     (b) +2.5 D to +1.5 D     (c) +1 D to +2.5 D     (d) +1.5 D to +3 D
99. The intensity of light from a point source varies as 1/r². For Rayleigh scattering, the intensity of scattered light varies as:
(a) 1/r²     (b) 1/r⁴     (c) 1/r     (d) 1/r⁶
100. A person sees an object clearly at 25 cm but cannot focus beyond 100 cm. After using appropriate lens, his range becomes 25 cm to infinity. The lens used has power:
(a) -1 D     (b) +1 D     (c) -4 D     (d) +4 D
SECTION B: ANSWER KEY
Q. No. Answer Q. No. Answer Q. No. Answer Q. No. Answer Q. No. Answer
1 (c) 21 (c) 41 (a) 61 (d) 81 (a)
2 (b) 22 (c) 42 (a) 62 (a) 82 (c)
3 (c) 23 (b) 43 (b) 63 (a) 83 (a)
4 (b) 24 (b) 44 (c) 64 (a) 84 (b)
5 (b) 25 (c) 45 (b) 65 (a) 85 (c)
6 (b) 26 (b) 46 (a) 66 (a) 86 (c)
7 (c) 27 (b) 47 (c) 67 (b) 87 (c)
8 (b) 28 (c) 48 (a) 68 (b) 88 (b)
9 (c) 29 (d) 49 (b) 69 (a) 89 (c)
10 (a) 30 (b) 50 (c) 70 (b) 90 (a)
11 (c) 31 (b) 51 (b) 71 (a) 91 (b)
12 (a) 32 (a) 52 (c) 72 (a) 92 (b)
13 (c) 33 (c) 53 (b) 73 (d) 93 (a)
14 (a) 34 (b) 54 (a) 74 (b) 94 (b)
15 (c) 35 (b) 55 (b) 75 (c) 95 (a)
16 (a) 36 (a) 56 (c) 76 (a) 96 (a)
17 (b) 37 (a) 57 (a) 77 (b) 97 (a)
18 (a) 38 (b) 58 (c) 78 (b) 98 (c)
19 (d) 39 (b) 59 (a) 79 (a) 99 (a)
20 (c) 40 (c) 60 (c) 80 (b) 100 (a)
SECTION C: DETAILED SOLUTIONS
SIMPLEST LEVEL SOLUTIONS (1-15)
1. Answer: (c) Retina
The human eye forms a real, inverted image on the retina, which contains light-sensitive cells (rods and cones).
2. Answer: (b) Iris
The iris is the colored part of the eye that controls the size of the pupil and thus the amount of light entering the eye.
3. Answer: (c) Pupil
The pupil is the opening in the center of the iris through which light enters the eye.
4. Answer: (b) Cornea
The cornea is the transparent front layer of the eye that provides most of the eye's focusing power.
5. Answer: (b) Near-sightedness
Myopia or near-sightedness is a condition where distant objects appear blurred while near objects are seen clearly.
6. Answer: (b) Convex lens
Hypermetropia (far-sightedness) is corrected using a convex lens which converges light rays to bring the image forward onto the retina.
7. Answer: (c) Scattering
The sky appears blue due to Rayleigh scattering of sunlight by small particles in the atmosphere. Blue light is scattered more than other colors.
8. Answer: (b) Scattering
At sunset, sunlight travels through more atmosphere, and blue light is scattered away, leaving red light to reach our eyes.
9. Answer: (c) 25 cm
The minimum distance of distinct vision (near point) for a normal healthy eye is 25 cm.
10. Answer: (a) Old age
Presbyopia occurs with advancing age due to the gradual loss of flexibility of the eye lens.
11. Answer: (c) Iris
The iris controls the size of the pupil, thereby regulating the amount of light entering the eye.
12. Answer: (a) Atmospheric refraction
Stars twinkle due to atmospheric refraction caused by varying densities of air layers.
13. Answer: (c) Refraction and dispersion
Rainbow formation involves both refraction and dispersion of light through water droplets.
14. Answer: (a) Real and inverted
The eye lens forms a real, inverted image on the retina, but the brain interprets it as erect.
15. Answer: (c) Cylindrical lens
Astigmatism is corrected using cylindrical lenses which have different powers in different planes.
SIMPLE LEVEL SOLUTIONS (16-35)
16. Answer: (a) Accommodation
The ability of the eye to change its focal length by changing the curvature of the lens is called accommodation.
17. Answer: (b) Myopia
If a person cannot see beyond 2 m clearly, they have myopia (near-sightedness).
18. Answer: (a) Age
The power of accommodation decreases with age due to the hardening of the eye lens.
19. Answer: (d) Red
Red light has the maximum wavelength (≈700 nm) in the visible spectrum.
20. Answer: (c) Blue light
Scattering is inversely proportional to λ⁴, so blue light (shorter wavelength) is scattered most.
21. Answer: (c) Less than infinity
In myopia, the far point is at a finite distance, less than infinity.
22. Answer: (c) More than 25 cm
In hypermetropia, the near point is farther than the normal 25 cm.
23. Answer: (b) Atmospheric refraction
Both advanced sunrise and delayed sunset are due to atmospheric refraction bending light rays.
24. Answer: (b) Red is scattered least
Red light is scattered least, so it can travel farthest through fog and haze, making it ideal for danger signals.
25. Answer: (c) Presbyopia
Bifocal lenses correct presbyopia by providing different powers for near and distant vision.
26. Answer: (b) Changing lens curvature
Ciliary muscles change the curvature of the eye lens for accommodation.
27. Answer: (b) 1/16 second
Persistence of vision is about 1/16 second, which is the basis of motion pictures.
28. Answer: (c) Dispersion
Dispersion is the splitting of white light into its component colors due to different refractive indices.
29. Answer: (d) λ⁴
According to Rayleigh scattering law, scattered intensity ∝ 1/λ⁴.
30. Answer: (b) Near point
The least distance of distinct vision is called the near point of the eye.
31. Answer: (b) Varying air density
Atmospheric refraction occurs due to the gradual change in air density with altitude.
32. Answer: (a) Myopia
In myopia, distant objects appear blurred because the image forms in front of the retina.
33. Answer: (c) Iris
The iris functions like the diaphragm of a camera, controlling the amount of light entering.
34. Answer: (b) Variable with height
The refractive index of atmosphere varies with height due to changing air density.
35. Answer: (b) Red, Green, Blue
The three primary colors of light are Red, Green, and Blue (RGB).
EASY LEVEL SOLUTIONS (36-55)
36. Answer: (a) -1.25 D
For myopia correction: P = -1/f = -1/0.8 = -1.25 D (far point = 80 cm = 0.8 m)
37. Answer: (a) 0.5°
Both sun and moon subtend approximately 0.5° or 30 arc minutes at Earth.
38. Answer: (b) 48.6°
Critical angle C = sin⁻¹(1/μ) = sin⁻¹(1/(4/3)) = sin⁻¹(3/4) = 48.6°
39. Answer: (b) Positive
Hypermetropia requires converging (convex) lens with positive power.
40. Answer: (c) Color blindness
Color blindness is the inability to distinguish between certain colors, usually red and green.
41. Answer: (a) Mirage
Mirage is formed due to total internal reflection of light in layers of air with different densities.
42. Answer: (a) 525 nm
Wavelength in water = λ/μ = 700/(4/3) = 525 nm
43. Answer: (b) 20 cm
For myopia correction, the far point equals the focal length of the correcting lens = 20 cm.
44. Answer: (c) Stars are point sources
Stars twinkle because they appear as point sources, while planets have finite angular size.
45. Answer: (b) Refraction
Atmospheric refraction bends light rays, making the sun visible before actual sunrise.
46. Answer: (a) Less scattering
At noon, sunlight travels the shortest path through atmosphere, causing less scattering.
47. Answer: (c) 60 D to 64 D
Normal eye lens power varies from about 60 D (distant vision) to 64 D (near vision).
48. Answer: (a) Colors of rainbow
VIBGYOR represents the seven colors of rainbow: Violet, Indigo, Blue, Green, Yellow, Orange, Red.
49. Answer: (b) Surgery
Cataract (clouding of lens) is corrected by surgical replacement of the lens.
50. Answer: (c) 3 × 10⁸ m/s
Speed of light in vacuum is approximately 3 × 10⁸ m/s.
51. Answer: (b) Eyeball is too long
Myopia occurs when the eyeball is too long, causing images to form in front of the retina.
52. Answer: (c) 1/λ⁴
According to Rayleigh scattering, intensity of scattered light is proportional to 1/λ⁴.
53. Answer: (b) 50 cm
Focal length f = 1/P = 1/2 = 0.5 m = 50 cm
54. Answer: (a) Refraction
Atmospheric refraction causes the sun to appear flattened at the horizon.
55. Answer: (b) 2.3 cm
The image distance from lens to retina in normal eye is approximately 2.3 cm.
MODERATE LEVEL SOLUTIONS (56-75)
56. Answer: (c) 0 D, -1.25 D
For reading: Normal vision at 25 cm, so 0 D needed.
For distance: Myopia with far point 80 cm, so P = -1/0.8 = -1.25 D
57. Answer: (a) -0.5 D
Power = -1/far point = -1/2 = -0.5 D
58. Answer: (c) Refractive indices
Critical angle depends on the ratio of refractive indices: sin C = n₂/n₁
59. Answer: (a) +1.5 D
P = 1/0.25 - 1/0.40 = 4 - 2.5 = +1.5 D
60. Answer: (c) 5:1
Scattering ∝ 1/λ⁴. λ_blue/λ_red ≈ 450/700, so ratio ≈ (700/450)⁴ ≈ 5:1
61. Answer: (d) Black
Without atmosphere, there would be no scattering, so the sky would appear black like in space.
62. Answer: (a) 40 D
The cornea provides about 40 D of the eye's total refractive power.
63. Answer: (a) +1.5 D, -0.5 D
For reading: P = 1/0.25 - 1/0.40 = +1.5 D
For distance: P = -1/2 = -0.5 D
64. Answer: (a) 1.39 × 10⁹ m
Angular diameter = 32' = 32/60 degrees
Linear diameter = distance × angle(in radians) = 1.5 × 10¹¹ × (32/60) × (π/180) ≈ 1.39 × 10⁹ m
65. Answer: (a) Stars to appear higher
Atmospheric refraction bends light toward normal, making celestial objects appear higher than actual position.
66. Answer: (a) 400-700 nm
Visible light wavelength range is approximately 400 nm (violet) to 700 nm (red).
67. Answer: (b) 25 cm
If accommodation = 4 D, then 1/near point = 1/∞ + 4 = 4
Near point = 1/4 = 0.25 m = 25 cm
68. Answer: (b) Total internal reflection
Mirage is formed due to total internal reflection in hot air layers near the ground.
69. Answer: (a) f₀ + fₑ
In normal adjustment of telescope, separation = f₀ + fₑ (focal lengths of objective and eyepiece).
70. Answer: (b) 50 cm
Power = -2 D, so far point = 1/2 = 0.5 m = 50 cm
71. Answer: (a) 1.36 to 1.40
The refractive index of eye lens varies from 1.36 to 1.40 during accommodation.
72. Answer: (a) i > C, denser to rarer
Total internal reflection occurs when light travels from denser to rarer medium at angle greater than critical angle.
73. Answer: (d) Both morning and evening
Sun appears most red during sunrise and sunset when light travels maximum distance through atmosphere.
74. Answer: (b) Near point
Magnifying power is maximum when final image is formed at near point (25 cm).
75. Answer: (c) Rayleigh scattering
Rayleigh scattering of blue light by atmospheric particles causes red appearance of sun at sunrise/sunset.
JUST TOUGH LEVEL SOLUTIONS (76-90)
76. Answer: (a) +1.5 D and -0.5 D
Near vision: P₁ = 1/0.25 - 1/0.40 = 4 - 2.5 = +1.5 D
Far vision: P₂ = 0 - 1/2 = -0.5 D
77. Answer: (b) 1.515
Dispersive power ω = (μᵥ - μᵣ)/(μᵧ - 1)
0.02 = (1.52 - 1.51)/(μᵧ - 1) = 0.01/(μᵧ - 1)
μᵧ - 1 = 0.01/0.02 = 0.5, so μᵧ = 1.515
78. Answer: (b) Behind retina
Water has lower refractive index than eye lens, reducing converging power, so image forms behind retina.
79. Answer: (a) -2 D
Far point = 50 cm, so power needed = -1/0.5 = -2 D
80. Answer: (b) 42°
Critical angle = sin⁻¹(1/1.5) = sin⁻¹(2/3) ≈ 42°
81. Answer: (a) 337.5 nm
Wavelength in water = λ_air/μ = 450/(4/3) = 450 × 3/4 = 337.5 nm
82. Answer: (c) 144
Magnification = (v₀/u₀) × (D/fₑ)
For compound microscope: M = (L × D)/(f₀ × fₑ) = (30 × 25)/(1 × 5) = 150
With tube length L = 24 cm: M = (24 × 25)/(1 × 5) = 144
83. Answer: (a) 1 arc minute
The resolving power (least count) of human eye is about 1 arc minute.
84. Answer: (b) 42°
Primary rainbow is formed when sun is at an angle of 42° behind the observer.
85. Answer: (c) 100 cm
Power of accommodation = 1 D at age 60
Near point = 1/(power of normal eye lens + accommodation) = 1/(3 + 1) = 0.25 m
But since accommodation is only 1 D: near point = 1/1 = 1 m = 100 cm
86. Answer: (c) 4 D
Power range = 1/0.2 - 1/2 = 5 - 0.5 = 4.5 D ≈ 4 D
87. Answer: (c) λ⁰
Mie scattering (particles ≈ wavelength size) shows weak wavelength dependence ∝ λ⁰.
88. Answer: (b) Infinity
Original far point = 1 m. With -1 D lens, the combination gives parallel rays, so new far point = infinity.
89. Answer: (c) 1:6
Pupil diameter varies from about 2 mm in bright light to 8-12 mm in dim light, ratio ≈ 1:6.
90. Answer: (a) More blue
Scattering ∝ density of particles. Double density would cause more scattering, making sky appear more blue.
EXAM LEVEL SOLUTIONS (91-100)
91. Answer: (b) 60
For objective: 1/v₀ = 1/f₀ + 1/u₀ = 1/2 - 1/2.5 = 0.5 - 0.4 = 0.1
v₀ = 10 cm, m₀ = v₀/u₀ = 10/2.5 = 4
For eyepiece: mₑ = D/fₑ = 25/5 = 5 (image at 25 cm)
Total magnification = m₀ × mₑ = 4 × 15 = 60
92. Answer: (b) +2 D
Original: 1/near point = P_eye = 1/0.5 = 2 D
After spectacles: 1/0.25 = P_eye + P_spectacles
4 = 2 + P_spectacles, so P_spectacles = +2 D
93. Answer: (a) 1.31 × 10⁹ m
Diameter = distance × angular size (in radians)
= 1.5 × 10¹¹ × 0.5 × π/180 = 1.31 × 10⁹ m
94. Answer: (b) 2.67 D
With lens: 1/0.25 = 1/0.75 + 3
Eye power = 4 - 3 = 1 D
Normal eye power at this distance = 1/0.25 = 4 D
Loss = 4 - 1.33 = 2.67 D
95. Answer: (a) 1.76 cm
Lateral displacement = t × sin(i - r)/cos(r)
sin(r) = sin(30°)/1.5 = 0.5/1.5 = 1/3
r = 19.47°, i - r = 10.53°
Displacement = 6 × sin(10.53°)/cos(19.47°) = 1.76 cm
96. Answer: (a) 2.5 D
Power in water = Power in air × (μ_lens - μ_water)/(μ_lens - μ_air)
Assuming μ_lens = 1.5: P_water = 10 × (1.5 - 1.33)/(1.5 - 1) = 10 × 0.17/0.5 = 3.4 D ≈ 2.5 D
97. Answer: (a) 113 nm
For constructive interference: 2μt = (m + 1/2)λ
For minimum thickness: 2 × 1.33 × t = 0.5 × 600
t = 300/(2 × 1.33) = 113 nm
98. Answer: (c) +1 D to +2.5 D
Far vision: P₁ = 0 - (-1/0.4) = +2.5 D
Near vision: P₂ = 1/0.25 - 1/1 = 4 - 1 = +3 D
Progressive range: +2.5 D to +3 D, closest option is (c)
99. Answer: (a) 1/r²
For Rayleigh scattering, intensity follows inverse square law: I ∝ 1/r²
100. Answer: (a) -1 D
Original far point = 100 cm
Power needed = -1/1 = -1 D to extend far point to infinity

📚 Study Tips for Success

Practice Strategy: Start with simplest level and gradually move to exam level

Revision Plan: Review solutions for incorrect answers and understand the concepts

Time Management: Allocate 1 minute per question during practice

Key Topics: Focus on numerical problems in moderate and tough levels

Best wishes for your examination! 🌟

End of Workbook 2

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