question_answer

The angle of minimum deviation for a thin prism with respect to air and when dipped in water will be:\[\left( _{a}{{\mu }_{g}}=\frac{3}{2},\,{{\,}_{a}}{{\mu }_{w}}=\frac{4}{3} \right)\]

A) \[\frac{1}{4}\]                                   

B) \[\frac{1}{8}\]

C)  \[\frac{1}{3}\]                                  

D)  \[\frac{1}{2}\]

Correct Answer: A

Solution :

For a prism of refractive index\[\mu \], angle of minimum deviation is given by                 \[\mu =\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \frac{A}{2}}\] When prism is thin, \[{{\delta }_{m}}\] is small and                 \[\frac{\sin A+{{\delta }_{m}}}{2}=\frac{A+{{\delta }_{m}}}{2}\]and\[\sin \frac{A}{2}=\frac{A}{2}\]                 \[\mu =\frac{(A+{{\delta }_{m}})/2}{A/2}\] \[\Rightarrow \]               \[{{\delta }_{m}}=(\mu -1)A\] When in air,                 \[{{\delta }_{m}}={{(}_{a}}{{\mu }_{g}}-1)A\]                 \[{{\delta }_{m}}=\left( \frac{3}{2}-1 \right)A=\frac{A}{2}\]                          ? (i) When dipped in water                 \[\delta {{'}_{m}}={{(}_{w}}{{\mu }_{g}}-1)A\]                 \[=\left( \frac{_{a}{{\mu }_{g}}}{_{a}{{\mu }_{w}}}-1 \right)A\]                 \[\delta {{'}_{m}}=\left( \frac{3/2}{4/3}-1 \right)A=\frac{A}{8}\]                ? (ii) Dividing Eq. (i) by Eq. (ii), we get                 \[\frac{\delta {{'}_{m}}}{{{\delta }_{m}}}=\frac{1}{4}\]