IIT JEE Advanced 2010 Paper 1

Using Snell’s law at face AB of the prism ABCD, we have
sin 60° = √3sinr ⇒ sin r =

1

2

⇒ r = 30°

The ray PQ is incident at an angle 45° on the face CD. For the total internal reflection at the face CD, we have
i ≥ iC
That is, sin i ≥ siniC
Here i = 45° and siniC = 1/µ Therefore,
sin 45° ≥ (√3)−1 ⇒

1

√2

≥

1

√3

Here,

1

√3

> √2 is true. Hence, the ray PQ gets totally internally reflected at the face CD.
To test the total internal reflection of ray QR at the face AD:
i ≥ iC
That is,
sin i ≥ siniC sin 30° ≥

1

√3

⇒

1

2

≥

1

√3

Here, √3 ≥ 2 is false, that is, the ray QR does not undergo total internal reflection at face AD; instead, it comes out in air.
Applying Snell’s law at point R, we get
√3sin30° = sin r' ⇒ sin r' =

√3

2

⇒ r' = 60°
The total deviation between the incident and emergent rays is
δ = δ1+δ2+δ3+δ4
= - 30° + 90° + 30°
= 90°
(Note: The clockwise deviation is taken as positive and anticlockwise deviation is taken as negative.)