16. A plano-convex lens is silvered at the plane surface. If radius of curvature is R and refractive index is n, the radius of curvature of the convex mirror formed is:
[IOM 07]
R/(n-1)
R/n
R(n-1)
Rn
(a) Effective focal length of silvered lens: 1/F = 2(n-1)/R ⇒ Radius of curvature = 2F = R/(n-1).
17. The distance between an object and a diverging lens is 'p' times the focal length. The lateral magnification 'm' is:
[IOM 06]
1/(1 + p)
1/(1 - p)
1
(1 + p)
(a) For diverging lens: m = f/(f + u) = 1/(1 + p) (since u = -pf).
18. The focal length of a lens in air is 30 cm. In water (μw = 1.33, μlens = 1.5), its focal length is:
[IOM 02]
120 cm
20 cm
90 cm
60 cm
(a) Using lensmaker's formula in water: fwater = fair × (μlens - 1)/(μlens/μw - 1) ≈ 120 cm.
19. Two lenses of power +12D and -2D are placed in contact. The focal length of the combination is:
[MOE 06]
10 cm
12.5 cm
16.6 cm
8.33 cm
(a) Pcomb = +10D ⇒ f = 1/P = 0.1 m = 10 cm.
20. The effective power if lenses of focal length +10 cm and -20 cm are combined is:
[MOE 2056]
+5D
-5D
+3D
-3D
(a) Pcomb = 1/0.1 + (-1/0.2) = +5D.
21. If an object is placed at the focus of a convex lens, the refracted rays are:
[MOE]
Converging
Diverging
Parallel
Both converging and diverging
(c) At focus, refracted rays become parallel to the principal axis.
22. When a convex lens (f = 12 cm) is immersed in water, its focal length becomes:
32. Two lenses (P = +1.75D and -1.25D) are combined. The focal length of the combination is:
[BPKIHS-97]
25 cm
10 cm
200 cm
50 cm
(c) Pcomb = +0.5D ⇒ f = 1/P = 2 m = 200 cm.
33. Two identical plano-convex lenses (f = 40 cm) are pressed together. To obtain a real, inverted image with magnification unity, the object distance is:
[BPKIHS]
80 cm
40 cm
20 cm
160 cm
(a) Combined focal length F = f/2 = 20 cm. For m = -1, u = 2F = 40 cm (for single lens). For two lenses, u = 80 cm.
34. The focal length of a convex lens is f. An object is placed at distance x from its first focal point. The ratio of image size to object size is:
[MOE]
f/x
x/f
f²/x²
x²/f²
(d) Transverse magnification m = f/(f + x) ≈ x/f (for small x).
35. A convex lens produces a real image n times the size of the object. The object distance is:
[MOE]
(n - 1)f
(n + 1)f
(n - 1)f/n
(n + 1)f/n
(b) m = -n = v/u ⇒ v = -nu. Using lens formula: u = (n + 1)f/n.
36. A convex lens produces a virtual image n times the size of the object. The object distance is:
[MOE]
(n - 1)f
(n + 1)f
(n - 1)f/n
(n + 1)f/n
(c) m = +n = v/u ⇒ v = nu. Using lens formula: u = (n - 1)f/n.
37. A concave lens of focal length f produces an image 1/n times the size of the object. The object distance is:
[MOE]
(n - 1)f
(n + 1)f
(n - 1)f/n
(n + 1)f/n
(a) m = +1/n = v/u ⇒ v = u/n. Using lens formula: u = (n - 1)f.
38. A plano-convex lens (μ, radius R) is silvered on the plane side. The system behaves like a concave mirror of radius:
39. The distance between a convex lens and a plane mirror is 10 cm. Parallel rays incident on the lens form an image at the optical center after reflection. The focal length of the lens is:
[MOE]
10 cm
20 cm
40 cm
5 cm
(a) For image to form at optical center, the lens-mirror distance must equal the focal length (f = 10 cm).
40. A convex lens (f = 20 cm) and concave lens (f = -5 cm) are coaxial. A parallel beam leaves as a parallel beam. The separation between lenses is:
[MOE]
5 cm
10 cm
15 cm
20 cm
(c) Separation d = f₁ - f₂ = 20 - 5 = 15 cm.
41. A lens (focal length f, aperture diameter d) forms an image of intensity I. If the central part (d/2 diameter) is blocked, the new intensity is:
[IOM 2017]
I/4
3I/4
I/2
I/16
(b) Blocking central d/2 reduces area by 1/4 ⇒ Intensity becomes 3I/4.
42. An object is placed 20 cm from a convex lens (f = 10 cm). The image is formed at:
[MOE]
20 cm on the same side
20 cm on the other side
3 cm on the same side
2 cm on the other side
(b) Using lens formula: 1/v = 1/f - 1/u ⇒ v = +20 cm (real image on the other side).
43. Two thin lenses (f₁, f₂) are placed at distance 'd'. For zero power, the separation 'd' is:
[MOE]
f₁ - f₂
f₁ + f₂
f₁/f₂
√(f₁f₂)
(b) Condition for zero power: d = f₁ + f₂.
44. A convex lens (f₁) and concave lens (f₂) in contact act as a convergent lens if:
49. For a convex lens (fv, fr) and concave lens (Fv, Fr):
[MOE]
fv < fr and Fv > Fr
fv < fr and Fv < Fr
fv > fr and Fv > Fr
fv > fr and Fv < Fr
(a) For convex: fv < fr (violet bends more). For concave: Fv > Fr (opposite behavior).
50. An equiconvex lens (f = 0.1 m) is cut into two equal parts perpendicular to the axis. The ratio of new focal lengths is:
[MOE]
1:1
1:2
2:1
1:4
(a) Each plano-convex part has focal length 2f ⇒ ratio 1:1.
51. A symmetric double convex lens (P = 4D) is cut into two equal parts. The power of each part is:
[MOE]
2D
3D
4D
8D
(a) Each plano-convex part has power ≈ 2D (exact calculation depends on surface curvature).
52. The focal length of a plano-convex lens equals the radius of curvature of its curved surface. The refractive index is:
[MOE]
1.3
1.5
1.6
1.8
(b) 1/f = (μ - 1)/R. Given f = R ⇒ μ = 1.5.
53. Rays from a luminous object focus at point A. A convex lens (f = 30 cm) is placed 30 cm from A. The new focus is at B. The distance AB is:
[MOE]
45 cm
15 cm
30 cm
60 cm
(a) Lens shifts focus by f ⇒ AB = 30 + 15 = 45 cm.
54. A convex lens forms a 4 cm image on a screen. When shifted, it forms a 16 cm image. The object length is:
[MOE]
64 cm
8 cm
10 cm
6 cm
(b) For two positions: object size = √(h₁h₂) = √(4 × 16) = 8 cm.
55. A convex lens forms images with magnifications 2 and 0.5 for two positions separated by 30 cm. Its focal length is:
[MOE]
20 cm
10 cm
30 cm
60 cm
(a) Distance between positions: d = 4f ⇒ f = 30/3 = 10 cm (Note: Correction—d = 4f ⇒ f = 30/4 = 7.5 cm).
56. A lens forms a real image on a screen 100 cm from the object. When moved 20 cm, another image forms. The focal length is:
[MOE]
12 cm
21 cm
24 cm
46 cm
(b) Using displacement method: f = (D² - d²)/4D = (100² - 20²)/400 ≈ 21 cm.
57. For a convex lens, maximum power occurs when:
[MOE]
R₁ = 10 cm, R₂ = ∞
R₁ = ∞, R₂ = 10 cm
R₁ = R₂ = 10 cm
R₁ = R₂ = 5 cm
(d) Power ∝ (1/R₁ - 1/R₂). Maximum when R₁ = R₂ = smallest value (5 cm).
58. An object (1.5 cm) is placed on the axis of a convex lens (f = 25 cm). A real image forms at 75 cm. The image height is:
[MOE]
4.5 cm
3.0 cm
0.75 cm
0.5 cm
(b) m = v/u = 75/37.5 = 2 ⇒ h' = 1.5 × 2 = 3 cm.
59. A concavo-convex lens (R₁ = 40 cm, R₂ = 20 cm, μ = 1.5) has focal length:
[MOE]
40 cm
-80 cm
80 cm
-40 cm
(c) 1/f = (1.5 - 1)(1/40 - 1/20) ⇒ f = 80 cm.
60. An aeroplane with a camera (f = 5 cm) photographs 5 km terrain on 5 cm film. The flying height is:
[MOE]
1 km
2 km
3 km
4 km
(a) m = v/u ⇒ 5 cm/5 km = 0.05 m/u ⇒ u ≈ 1 km.
61. An aeroplane is flying at a height of 1500m
. It has a camera having convex lens of
focal length 45 cm with photographic plate
30cm x 30cm. How much area on the
ground can be photographed at one time ?
10'm
10'm?
c. 10'm2"
10'm
(b) Height of aeroplane H = 1500 m. Focal length of lens f = 45 cm = 0.45 m. Size of photographic plate = 30 cm x 30 cm = 0.3 m x 0.3 m. Using similar triangles, height of image / height of object = focal length / distance of object. Size of image / Area on ground = f / H. Area of image = 0.3 * 0.3 = 0.09 m². 0.09 / Area on ground = 0.45 / 1500. Area on ground = 0.09 * 1500 / 0.45 = 9 * 1500 / 45 = 1500 / 5 = 300 m². There seems to be error in options. Let's recheck. Area on ground = (0.3/0.45)² * 1500² = (2/3)² * 225 * 10⁴ = 4/9 * 225 * 10⁴ = 100 * 10⁴ = 10⁶ m². Option b has 10⁶ m³ which is likely a typo.
62. A cyclist is moving perpendicular to
principal axis at a distance of 10m with a
speed of 10m/s infront of a convex lens of
focal length 10cm. Find the time of
exposure of the lens if the image
displaced by 1mm on the photographi
plate.
a. 25 sec
50 sec
100 sec .
(d) Object distance u = 10 m = 1000 cm. Focal length f = 10 cm. Using lens formula, 1/v - 1/u = 1/f => 1/v - 1/(-1000) = 1/10 => 1/v = 1/10 - 1/1000 = 99/1000 => v = 1000/99 ≈ 10.1 cm. Speed of object = 10 m/s. Speed of image vi / vo = (v/u)² => vi = vo * (v/u)² = 1000 cm/s * (10.1/1000)² ≈ 0.01 cm/s = 0.1 mm/s. Displacement of image = 1 mm. Time of exposure = Displacement / Speed = 1 mm / 0.1 mm/s = 10 s. None of the options match.
63. A plane convex lens has diameter 6cm and
thickness from the centre is 3mm. If the
speed of light in the lens is 2x 10 m/s, then
the focal length of plane convex lens is
a. 10cm
b.
15cm
c. 20cm
d. 30cm
(c) Refractive index of lens μ = speed of light in vacuum / speed of light in lens = 3 × 10⁸ / 2 × 10⁸ = 1.5. Using lens maker's formula 1/f = (μ - 1) (1/R₁ - 1/R₂). For plane convex lens, R₂ = ∞. Radius of curvature of convex surface R₁ = (r² + t²) / 2t = (3² + 0.3²) / (2 * 0.3) = (9 + 0.09) / 0.6 = 9.09 / 0.6 ≈ 15.15 cm. 1/f = (1.5 - 1) (1/15.15 - 1/∞) = 0.5 * (1/15.15) => f = 15.15 / 0.5 ≈ 30.3 cm. Closest option is 30 cm.
64. A picture of size 2cm x 4em is shown on a
projector. If the magnification produced
be 10, the area of the image on the screen
will be
80cm
b. 800cm
d. 640cm
c. 8cm
(b) Magnification m = size of image / size of object = 10. Area of object = 2 * 4 = 8 cm². Area of image / Area of object = m². Area of image / 8 = 10² = 100. Area of image = 8 * 100 = 800 cm².
65. If lens behaves as converging in air and
diverging in water. Then refractive index is
[TOM 20151
[TOM 20151]
Less than 1.33
Between 1 to.1.33
More than 1.33 .
33
(a) Lens maker formula 1/f = (μlens/μmedium - 1) (1/R₁ - 1/R₂). For converging lens in air, f > 0 => μlens > μair (1). For diverging lens in water, f < 0 => μlens < μwater (1.33). So, refractive index of lens is between 1 and 1.33.
66. When the convex lens of refractive index
(H), immersed in water of same refractive
index (1) then, its focal length is:
[KU 2016]
a.
0
d. Increases.
c. decreases
(b) Using lens maker's formula 1/f = (μlens/μmedium - 1) (1/R₁ - 1/R₂). If μlens = μmedium = μ, then 1/f = (μ/μ - 1) (1/R₁ - 1/R₂) = (1 - 1) (1/R₁ - 1/R₂) = 0. So, f = ∞.
67. A lens made of glass of refractive index
1.52 has focal length of 10 cm in air and 50
cm when immersed in liquid. The
refractive index of liquid must be:
[KU 2017]
68. The focal length of lens is F, and diameter
of aperture is d. When y of diameter is
blackened , then intensity of image will be:
[IOM 2017]
[IOM 2017]
b.
Al-
151
16
d.
16
c.
16
() Intensity of image is proportional to the area of the aperture. Initial area A₁ = π(d/2)². When y of diameter is blackened, the remaining diameter is d - y. Remaining radius = (d - y) / 2. Remaining area A₂ = π((d - y) / 2)². Intensity I₂ / I₁ = A₂ / A₁ = [π((d - y) / 2)²] / [π(d/2)²] = (d - y)² / d² = (1 - y/d)². Option a has Al-. Assuming it's I₁. If y = d/4 is blackened, I₂/I₁ = (1 - 1/4)² = (3/4)² = 9/16. Option b has 151/16.
