Optics MCQ with Answers

36) An object is placed at a distance of 10 cm from a co-axial combination of two lenses A and B in contact. The combination forms a real image three times the size of the object. If lens B is concave with a focal length of 30 cm, what is the nature and focal length of lens A ?

Answer is:

Convex, 6 cm

37) A plano-convex lens acts like a concave mirror of 28 cm focal length when its plane surface is silvered and like a concave mirror of 10 cm focal length when its curved surface is silvered. What is the refractive index of the material of the lens?

Answer is:

1.55

Explanation:

n1/u +n2/v=(n2-n1)/r for first refraction by convex surface: u??, v=rn2/(n2-n1) for plane reflection: r is infinite,v'=-v for second refraction: 1.5/v' -1/v''=(1-1.5)/r 1/v''+1/v=1/f power=1/f

38) A pin is placed 10 cm in front of a convex lens of focal length 20 cm and refractive index 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature of 22 cm. How far from the lens is the final image formed?

Answer is:

11 cm

Explanation:

The curved silvered surface will behave as a concave lens of focal length fm=R/2=?22/2 =?11cm =?0.11m And hence PM=the power of mirror =?1/fM=?1/?0.11=1/0.11D Further as focal length of lens is 20 cm ie 0.2m its power is P2=1/f2P2 =1/20D Now as in image formation, light after passing through the lens will be reflected back by curved mirror through lens again . P=PL+PM+PL =2PL+PM P=2/0.2+1/0.11=210/11D So, focal length of equivalent mirror. F=?1/P=11/210m ?110/21cm ie silvered lens behaves as a concave mirror of focal length (110/21)cm So far object at a distance 10 cm in front of it. 1/V+1/?10=21/110 i,e V=?11cm The image will be 11 cm in front of silvered lens and will be real. Hence c is the correct answer.

39) A telescope, when in normal adjustment, has a magnifying power of 6 and the objective and the eye-piece are 14 cm apart The focal lengths of the eye-piece and the objective respectively are

Answer is:

2 cm and 12 cm

Explanation:

The most commonly used formula in amateur astronomy is used to calculate the magnification of a telescope: magnification = focal length of telescope / focal length of eyepiece. Example: using a 10mm eyepiece in a telescope with a focal length of 1000mm results in a magnification of 100x (1000 / 10 = 100)

40) A telescope has an objective of focal length 100 cm and an eye-piece of focal length 5 cm. What is the magnifying power of the telescope when it is in normal adjustment?

Answer is:

20

Explanation:

The most commonly used formula in amateur astronomy is used to calculate the magnification of a telescope: magnification = focal length of telescope / focal length of eyepiece. Example: using a 10mm eyepiece in a telescope with a focal length of 1000mm results in a magnification of 100x (1000 / 10 = 100)