question_answer

A, B and C are the parallel sided transparent media of refractive indices\[{{n}_{1}}\],\[{{n}_{2}}\]and\[{{n}_{3}}\]respectively. They are arranged as shown in the figure. A ray is incident at an angle \[i\] on the surface of separation of A and B which is as shown in the figure. After the refraction into the medium B, the ray grazes the surface of separation of the madia B and C. Then, \[sin\text{ }i\]equals to

A)  \[\frac{{{n}_{3}}}{{{n}_{1}}}\]

B)  \[\frac{{{n}_{1}}}{{{n}_{3}}}\]

C)  \[\frac{{{n}_{2}}}{{{n}_{3}}}\]

D)  \[\frac{{{n}_{1}}}{{{n}_{2}}}\]

Correct Answer: A

Solution :

Applying Snells law between the surfaces A and B \[{{n}_{1}}\sin i={{n}_{2}}\sin {{r}_{1}}\] ?(i) Again applying Snells law between surfaces B and C \[{{n}_{2}}\sin \,{{r}_{1}}={{n}_{3}}\sin {{r}_{2}}\] ... (ii) From Eqs. (i) and (ii), we get \[{{n}_{2}}\sin \,i={{n}_{3}}\sin {{r}_{2}}\] Here,              \[{{r}_{2}}={{90}^{o}}\] \[\therefore \] \[{{n}_{1}}\sin i={{n}_{3}}\] \[\Rightarrow \] \[\sin i=\frac{{{n}_{3}}}{{{n}_{1}}}\]