question_answer

A thick transparent slab is made such that its refractive index changes from \[{{n}_{1}}\] to \[{{n}_{2}}\] nearly linearly. This slab is used by a student to measure lateral displacement in lab. She draw following diagram using four pins. Angle p must be

A) \[\alpha \]                                 

B) \[\frac{{{n}_{1}}}{{{n}_{2}}}\alpha \]

C) \[\frac{{{n}_{2}}}{{{n}_{1}}}\alpha \]

D)                    \[{{n}_{1}}{{n}_{2}}\alpha \]

Correct Answer: A

Solution :

We divide slab into a series of parallel slabs with different refractive indices.

We have, \[\frac{\sin \alpha }{\sin 1}=\frac{{{n}_{1}}}{{{n}_{0}}}\]

\[\frac{\sin 1}{\sin 2}=\frac{{{n}_{2}}}{{{n}_{1}}},\frac{\sin 2}{\sin 3}=\frac{{{n}_{3}}}{{{n}_{2}}}\]

\[\frac{\sin 3}{\sin 4}=\frac{{{n}_{4}}}{{{n}_{3}}},\frac{\sin 4}{\sin \beta }=\frac{{{n}_{0}}}{{{n}_{4}}}\]

So, we have

\[\frac{\sin \alpha }{\sin 1}\times \frac{\sin 1}{\sin 2}\times \frac{\sin 2}{\sin 3}\times \frac{\sin 3}{\sin 4}\times \frac{\sin 4}{\sin \beta }=1\]

\[\Rightarrow \]\[\sin \alpha =\sin \beta \]or \[\alpha =\beta \]