IIT JEE Advanced 2010 Paper 1

Using Snell’s law at face AB of the prism ABCD, we have sin 60° = $\sqrt{3} \sin r$ ⇒ sin r = $\frac{1}{2}$ ⇒ 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 ≥ $i_C$ That is, sin i ≥ $\sin i_C$ Here i = 45° and $\sin i_C$ = 1/µ Therefore, sin 45° ≥ $(\sqrt{3})^{-1}$ ⇒ $\frac{1}{\sqrt{2}}$ ≥ $\frac{1}{\sqrt{3}}$ Here, $\frac{1}{\sqrt{3}}$ > $\sqrt{2}$ 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 ≥ $i_C$ That is, sin i ≥ $\sin i_C$ sin 30° ≥ $\frac{1}{\sqrt{3}}$ ⇒ $\frac{1}{2}$ ≥ $\frac{1}{\sqrt{3}}$ Here, $\sqrt{3}$ ≥ 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 $\sqrt{3} \sin 30^{\circ}$ = sin r' ⇒ sin r' = $\frac{\sqrt{3}}{2}$ ⇒ r' = 60° The total deviation between the incident and emergent rays is δ = $\delta_1+\delta_2+\delta_3+\delta_4$ = - 30° + 90° + 30° = 90° (Note: The clockwise deviation is taken as positive and anticlockwise deviation is taken as negative.)