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

Total Questions: 50

31. A convex lens having power 5 D is placed in contact with a concave lens having power - 3 D. The focal length of the combination will be: [RRC Group D 22/08/2022 (Afternoon)]

Correct Answer: (a) 50 cm
Solution:

50 cm. Power of Convex Lens, ๐‘ƒโ‚ = 5 D. Power of a Concave lens, ๐‘ƒโ‚‚ = -3 D . The Combined Power, P = ๐‘ƒโ‚+ ๐‘ƒโ‚‚,
P = 5 +(-3) , P = 2 D . Since
P = 1/focal length , f = 1/p , f = 1/2
f = 0.5m = 50cm. Hence, the focal length of the combination is 50 cm.

32. A concave mirror having focal length of magnitude 20 cm forms a real image at a distance of 60 cm from it. The object distance (in cm) is [RRC Group D 22/08/2022 (Evening)]

Correct Answer: (c) -30
Solution:

- 30. The focal length of the concave mirror is always negative, Given f = -20 cm. The image distance of the concave mirror is negative (image is formed in front of the mirror (real image)), Given v = -60 cm.
Mirror formula :- 1/v + 1/u = 1/f

33. A convex lens produces a magnification of -3 for an object placed at 1.5 m from the lens. Find the image distance (with correct sign) [RRC Group D 22/08/2022 (Evening)]

Correct Answer: (b) 4.5 m
Solution:

4.5 m.
Object Distance, u = - 1.5m
(for convex lens), Magnification, m = - 3,
Image distance, v = ?.
Since, m = v/u.
So, (-3) = v/(-1.5)ย  โ‡’ (-3) ร— (-1.5) = v โ‡’ย  4.5.
Hence, Image distance (v)= 4.5 m.

34. A light ray is traveling from air medium to water medium (refractive index = 1.3) such that angle of incidence is x degree and angle of refraction is y degree. The value of ratio (sin y)/(sin x) is: [RRC Group D 24/08/2022 (Evening)]

Correct Answer: (b) 1/1.3
Solution:

1 /1.3 According to Law of Refraction (Snell's law), The ratio of the sine of the angle of incidence โ€˜iโ€™ to the sine of the angle of refraction โ€˜rโ€™ is constant for the pair of given media. This constant is called the refractive index of the second medium w.r.t. the first medium and can be expressed as: (sin i) / (sin r) = refractive index of 2nd medium / refractive index of first medium So, (sin x) / (sin y) = 1.3/1 โ‡’ (siny) / (sinx) = 1/1.3.

35. Identify the correct relation between radius of curvature 'R", object distance "u" and image distance "v" for a spherical [RRC Group D 24/08/2022 (Evening)]

Correct Answer: (b)
Solution:

It can be derived by using the
mirror formula,
1/f = 1/u + 1/vย  & R = 2f
Here, u = object distance from the mirror
v = image distance from the mirror
R = radius of curvature
f = focal length
Solving the equation and substituting f by
R, we get R = 2uv/u+v .
m = ๐ป๐‘’๐‘–๐‘”โ„Ž๐‘ก ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘–๐‘š๐‘Ž๐‘”e/๐ป๐‘’๐‘–๐‘”โ„Ž๐‘ก ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘œ๐‘๐‘—๐‘’๐‘t = - v/u

36. An object, 1.0 cm in height, is placed at a distance of 18.0 cm in front of a concave mirror of focal length 10.0 cm, on its principal axis. Its image has a height of_________and is__________. [RRC Group D 25/08/2022 (Morning)]

Correct Answer: (a) more than 1.0 cm, inverted
Solution:

more than 1.0 cm, inverted. Given, object height, hโ‚ = 1.0 cm, object distance, u = - 18.0 cm, & Focal length = - 10.0 cm.So, the image formed will be more than 1.0 cm, inverted (because hโ‚‚ has a negative sign).

37. The refractive indices of turpentine oil and glass are 1.47 and 1.52, respectively. A ray of light passes from turpentine oil to glass. The refractive index of glass with respect to turpentine oil is ________ and the ray bends ________ the normal in glass. [RRC Group D 25/08/2022 (Morning)]

Correct Answer: (c) 1.03, towards
Solution:

1.03, towards. The refractive index of glass with respect to turpentine oil = refractive index of glass / refractive index of turpentine oil = 1.52/1.47 = 1.03. Now, since the ray of light passes from turpentine oil to glass i.e. from a rare medium to a denser medium, it will bend towards the normal.

38. An object is placed on the principal axis of a concave lens of focal length 20 cm, at a distance of 10 cm. The magnification produced by the lens is and the image is: [RRC Group D 25/08/2022 (Morning)]

Correct Answer: (a) less than 1, erect
Solution:

less than 1, erect.
Given, Distance of the object, u = - 10 cm
& Focal length, f = - 20 cm
Using the Lens formula, =ย  1/๐‘“ = 1/๐‘ฃ - 1/๐‘ข
โ‡’ 1/v = 1/f + 1/u =ย  1/(-20) + 1/(-10)
โ‡’ v = - 20/3
Now, the magnification for a lense, m =
v/u = - (20/3)/(-10) = 2/3 = Less than 1.
Since the value of the magnification is positive, the image formed is erect.

39. An object, 3.0 cm in height, is placed at a distance of 20.0 cm in front of a convex mirror of focal length 6.0 cm on its principal axis. Its image has a height of ________ and is ________. [RRC Group D 25/08/2022 (Afternoon)]

Correct Answer: (d) less than 3.0 cm, erect
Solution:

less than 3.0 cm, erect.
according to sign conventions, when the magnification is positive then the image has to be erect and virtual. if it is negative then the image is real and inverted. Given, Object height, hโ‚’ = 3.0 cm,
object distance (u) = - 20.0 cm,
Focal length (f) = + 6.0 cm.Hence, the image height is less than 3.0 cm, erect.

40. A ray of light is incident at a point M on a convex mirror (pole P) of radius of curvature 20 cm. It is reflected back along the same path, and appears to come from its center of curvature C. Following the new Cartesian sign convention, PC = ________. [RRC Group D 26/08/2022 (Afternoon)]

Correct Answer: (c) 20 cm
Solution:

20 cm. Signs of the radius of curvature and focal length in convex mirrors are taken as; Radius of curvature = +ve, Focal length = +ve. And on reflecting back light ray will pass through the centre of curvature. So, PC = +20 cm.