Railway Science (Physics-Light and Optics) (Part-VII)

Total Questions: 50

41. A ray of light is traveling from medium A to medium B. The incident ray makes an angle 75° with respect to the normal, and the refracted ray makes an angle 40° with respect to the normal. The refractive index of the medium A relative to the medium B is: [RRC Group D 28/09/2022 (Morning)]

Correct Answer: (a) 0.67
Solution:

0.67. Refractive Index = 𝑠𝑖𝑛 r/𝑠𝑖𝑛 i (From medium A to B )
Refractive index = 𝑠𝑖𝑛 40/ 𝑠𝑖𝑛 75 =  0.67.

42. If a coin is placed at the bottom of a tumbler of height 30 cm filled with water of refractive index 1.33, then the apparent depth of the coin appears to be: [RRC Group D 28/09/2022 (Evening)]

Correct Answer: (d) 22.6 cm
Solution:

22.6 cm. Given, real depth = 30 cm, apparent depth = ? and n = 1.33
Refractive index (n) = 𝑟𝑒𝑎𝑙 𝑑𝑒𝑝𝑡ℎ/𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑑𝑒𝑝𝑡ℎ ,
⇒ 1.33 = 30/𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑑𝑒𝑝𝑡ℎ ,
⇒ apparent depth = 30/1/33 , ⇒ 22.6 cm.

43. A doublet is a combination of two lenses. One such doublet is made with a convex lens of focal length of 20 cm and a concave lens of focal length of 50 cm. The effective focal length and power of this doublet is ______and________ respectively. [RRC Group D 29/09/2022 (Morning)]

Correct Answer: (b) 33.3 cm, 3D
Solution:

44. A ray of light in air is incident at an angle of 60° on the surface separating air from a medium of refractive index √(3/2) The ray is refracted in the medium at an angle of___________. [RRC Group D 29/09/2022 (Afternoon)]

Correct Answer: (a) 45°
Solution:

45. A coin of diameter 4 cm is placed at a distance of 45 cm in front of a convex lens of focal length 30 cm. Then the diameter of the image of the coin will be: [RRC Group D 29/09/2022 (Evening)]

Correct Answer: (b) 8 cm
Solution:

8 cm. Given, Height of object (hₒ) = 4cm, Object distance (u) = - 45 cm, focal length (f) = 30 cm.
Formula used, Lens formula,

46. Suppose a satellite is 10² km above the ground and it is used to take high resolution images of objects. If a concave mirror is used to form a primary image of an object of size 1.0 m and if the image size is 5 μm and it is inverted, then the principal focus of the concave mirror should be: [RRC Group D 30/09/2022 (Afternoon)]

Correct Answer: (d) −0.5 m
Solution:Here, Object distance > Image distance, we can say that the object was at infinity with respect to image. So, in such cases the image will form on focus. So, focal length = image distance = - 0.5 m.

47. The refractive index of a diamond is 2.42. Then the speed of light in the diamond is: [RRC Group D 06/10/2022 (Evening)]

Correct Answer: (b) 1.24 × 10⁸ m/s
Solution:

We know that refractive index (n) = c/v = (speed of  light in vaccum)/(speed of light in diamond)
⇒ 2.42 = (3 × 10⁸)/v
⇒ v = 1.24 × 10⁸ m/s

48. Suppose Raghu has kept an object in front of a concave mirror of focal length (f) at various distances (u) and he has measured the corresponding image distances (v). From such an experiment, Raghu is able to plot a graph of u against 1/magnification, i.e. u vs. 1/m. Which of the following options is true? [RRC Group D 06/10/2022 (Evening)]

Correct Answer: (d) It is a straight line with slope +f, and x intercept −1 and y intercept +f
Solution:

In a concave mirror, when the distance of the object is less than the focal length, the magnification will be greater than one. When we place the object on the centre of curvature, magnification will be 1. When we move away from the mirror, magnification will be less than 1.

49. Ram has a corrective lens of power − 6.5 D. The focal length of the lens is: [RRC Group D 07/10/2022 (Evening)]

Correct Answer: (d) −15.38 cm
Solution:

The power of a lens is defined as the reciprocal of its focal length in meters.
Since P = 1/f  ⇒  f = 1/D = 1/-6.5
= - 0.1538 meters
= -0.1538 × 100 = - 15.38 cm.

50. If a lens has a focal length of 25 cm, what will be the power of that lens? [RRB NTPC CBT - I ( 10/02/2021 ) Evening]

Correct Answer: (a) 4D
Solution:

Power of lens (Dioptre)
D = 1/focal length(f)
F = 25 cm = 0.25m
D = 1/0.25m = 4D