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 \[\sin \frac{A}{2}=\frac{A}{2}\] \[\therefore \] \[\mu =\frac{(A+{{\delta }_{m}})/2}{A/2}\] \[\Rightarrow \] \[{{\delta }_{m}}-(\mu -1)A\] 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 \[\frac{\delta _{m}^{}}{{{\delta }_{m}}}=\frac{1}{4}\]