Today we will discuss two multiple choice questions on optics, which were included in the IIT-JEE 2010 question paper. The first question is *single correct choice* type mcq where as the second one is *multiple correct choice* type mcq.

(1) A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

(A) virtual and at a distance of 16 cm from the mirror

(B) real and at distance of 16 cm from the mirror

(C) virtual and at a distance of 20 cm from the mirror

(D) real and at a distance of 20 cm from the mirror

The lens has focal length *f* = 15 cm and the object O is placed at distance 30 cm which is equal to 2*f*. Therefore a real image I_{1} will be formed at distance 2*f* (= 30 cm) on the other side of the lens, if the mirror is absent.

Since the image I_{1} is at 20 cm from the plane mirror, a real image I_{2} would be formed (by the reflection of the rays at the plane mirror) at distance 20 cm from the mirror. But the returning rays are further refracted by the lens and the final image is formed at I_{3}. The image I_{2 }serves as the object for the lens for the formation of the final image I_{3}.

As is clear from the figure, the object distance *u*_{3} for the formation of the final image is 10 cm. Therefore we have

1/*v*_{3} – 1/10 = 1/15

[We have used the law of distances 1/*v* – 1/*u* = 1/*f*, in accordance with the Cartesian sign convention]

Therefore, 1/*v*_{3} = 1/15 + 1/10

This gives *v*_{3} = 6 cm.

The image is* real* and its distance from the plane mirror is (6+10) cm = 16 cm [Option (b)].

(2) A ray OP of monochromatic light is incident on the face AB of prism ABCD near the vertex B at an incident angle of 60º (see figure). If the refractive index of the material of the prism is √3 , which of the following is (are) correct ?

(A) The ray gets totally internally reflected at face CD

(B) The ray comes out through face AD

(C) The angle between the incident ray and the emergent ray is 90º

(D) The angle between the incident ray and the emergent ray is 120º

When the angle of incidence (at P) is 60º, the angle of refraction *r *is 30º since the refractive index *n *of the material of the prism is given by

*n =* sin *i/*sin *r *

Or, √3 = sin 60º/sin* r* so that sin *r =* (sin 60º)/√3 = ½ from which *r = *30º

From the quadrilateral PABQ (fig.) it follows that angle PQB = 45º. Therefore the angle of incidence of the ray at the point Q is 45º. But this is greater than the critical angle for the interface.

[If *θ*_{c} is the critical angle for the interface, we have 1/sin* θ*_{c} = *n =*√3 so that sin* θ*_{c} = 1/√3. Since sin* *45º = 1/√2 it follows that 45º >* θ*_{c}]

The ray incident at Q is therefore totally reflected and it is incident at an angle of 30º at the point R. The angle of refraction at R is 60º and hence the angle between the incident ray and the emergent ray is 90º.

Options A, B and C are therefore correct.

[If you know the action of the prism commonly used in the *constant deviation spectrograph*, you will be able to answer this question immediately].