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

Total Questions: 50

21. Suppose a ball is placed in front of a concave mirror and a real image that is twice the size of the ball is formed on a screen. The ball and the screen are then moved until the image is five times the size of the object. If the shift of the screen is d, then the shift in the object is: [RRC Group D 19/09/2022 (Afternoon)]

Correct Answer: (b) d/10
Solution:

22. A ball of diameter 10 cm is placed at a distance of 40 cm in front of a lens with a power of +5.0 D. The diameter of the image of the ball will be: [RRC Group D 19/09/2022 (Afternoon)]

Correct Answer: (a) 10 cm
Solution:

23. When an object is placed at a point 21 cm in front of a convex mirror, the image is formed at 7 cm behind the mirror. Now, someone moves the object to a distance of 14 cm in front of the mirror. The distance of the image from the mirror (in cm) now is: [RRC Group D 19/09/2022 (Evening)]

Correct Answer: (a) + 6
Solution:

24. In order to obtain an inverted and magnified image of an object by a spherical mirror of radius of curvature 40 cm, the object should be placed: [RRC Group D 20/09/2022 (Morning)]

Correct Answer: (a) between 20 cm and 40 cm from a concave mirror
Solution:Radius of curvature (R) = 40 cm.
f = R/2 = 40/2 = 20 cm
So, distance from the pole of the mirror = 20 cm When an object is placed in between the center of curvature and focus, the real image is formed behind the center of curvature which is inverted and the size of the image is larger than that of the object.From the above theorem it states, if the object is placed between the center of curvature and focus which is between 20 - 40 cm from the pole of the mirror and image is real and inverted.

25. The magnification of an image is +1.5 and the object distance is 30 cm from a spherical mirror. The image is formed at___________. [RRC Group D 20/09/2022 (Afternoon)]

Correct Answer: (c) 45 cm behind the mirror
Solution:45 cm behind the mirror. Given, magnification of an image (m) = +1.5. Object distance (u) = - 30 cm.
magnification (m) =  -v/u  ⇒ 1.5 =  -v/-30.
⇒1.5 × 30 = v ⇒ v = 45cm.
Since, magnification is positive, hence the image will form 45 cm behind the mirror. Image is erect and virtual.

26. A shaving mirror is constructed in such a way that a person at a distance of 30 cm from the mirror sees his image magnified 1.33 times. What will be the radius of curvature of the mirror? [RRC Group D 20/09/2022 (Afternoon)]

Correct Answer: (d) 34.25cm
Solution:

27. Suppose a pin is placed in front of a concave mirror and a real image that is thrice the size of the pin is formed on a screen. The pin and the screen are then moved until the image is six times the size of the object. If the shift of the screen is 24 cm, then the shift in the object is: [RRC Group D 20/09/2022 (Evening)]

Correct Answer: (b) 4/3 cm
Solution:

28. Suppose a dentist uses a spherical mirror which can result in an upright image that is magnified five times. Then the radius of curvature in terms of the object distance 𝑑。is given by: [RRC Group D 20/09/2022 (Evening)]

Correct Answer: (a)
Solution:

29. A light ray travels from air into an optical fibre with an index of refraction of 1.45. If the angle of incidence on the end of the fibre is 22°, the angle of refraction inside the fibre is: [RRC Group D 22/09/2022 (Afternoon)]

Correct Answer: (c) 14.99°
Solution:

30. Suppose a beam of light of average wavelength 600 nm is incident on a glass prism from air, and on entering into the prism, it splits into different colours. One of the colours has a wavelength 380 nm. The refractive index of the medium for this particular wavelength is: [RRC Group D 22/09/2022 (Evening)]

Correct Answer: (a) 1.58
Solution:

1.58. The refractive index of the material of the prism is given by μ=c/v,
where c is the speed of light in vacuum, and v is the speed of light in the medium (prism). Since the velocity of a wave is a product of frequency and wavelength, we can write c = vλₐ and v = vλₘ,where λₐ and λₘ are the wavelengths in air and medium respectively and v is the frequency of light waves.
Thus, μ= 𝑣λ𝑎/𝑣λm = λₐ/λₘ. For 380 nm wavelength, the refractive index is μ = 600/380 = 1.578 ≈ 1.58.