QJEEAdv10Phdbde: JEE Advanced 2010 Physics - Class 12 JEEAdv10Phdbde Solution: A ray OP of monochromatic light is incident on the face AB of…

Consider the refraction at the face AB. Snell's law,

$\sin i/\sin r = \sin 60^\circ /\sin r = \sqrt 3$ ,

gives $r = 30^\circ$ .

The geometry in BCQP gives

$\angle BPQ = 30^\circ + 90^\circ = 120^\circ$

$\angle CQP = 360^\circ - (135^\circ + 60^\circ + 120^\circ ) = 45^\circ$ .

Thus, the angle of incidence at Q is ${i_1} = 45^\circ$ . The critical angle for prism to air refraction is given by $\sin {i_c} = 1/\sqrt 3$ . Since \sin {i_1} = 1/\sqrt 2 > 1\sqrt 3 , we get {i_1} > {i_c} i.e., the angle of incidence is greater than the critical angle. Thus, the ray undergoes total internal reflection at Q. The laws of reflection gives ${r_1} = {i_1} = 45^\circ$ . In triangle QRD, $\angle QRD = 60^\circ$ and hence the angle of incidence at R is ${i_2} = 30^\circ$ .

Applying Snell's law at face AD, we get

$\sqrt 3 \times \sin 30^\circ = 1 \times \sin e$

or, $\sqrt 3 \times {1 \over 2} = \sin e$

$\sin e = {{\sqrt 3 } \over 2}$

or, $e = {\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) = 60^\circ$

From figure,

The angle between the incident ray and the emergent ray is 90 $^\circ$ .

Hence, option (c) is correct and option (d) is incorrect.

Note : Angle between incident and emergent rays is the same as the angle between the two faces = 90 $^\circ$ .