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 \[\frac{\delta {{'}_{m}}}{{{\delta }_{m}}}=\frac{1}{4}\]You need to login to perform this action.
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