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 \[\alpha =\beta \] |
You need to login to perform this action.
You will be redirected in
3 sec