Question

Derive an expression for the refractive index of the material of a prism.

Answer 1

$d=\angle OMN+\angle ONM$

$i_1=\angle OMN+r_1$

$\angle OMN=i_1-r_1$

$i_2=\angle ONM+r_2$

$\angle ONM=i_2-r_2$

$d=(i_1-r_1)+(i_2-r_2)$

$d=(i_1+i_2)-(r_1+r_2)$

$A+\angle PMN+\angle MNP={180}^0$

$\angle PMN={90}^0-r_1$

$\angle MNP={90}^0-r_2$

$A+({90}^0-r_1)+({90}^0-r_2)={180}^0$

$A=r_1+r_2$

$d=(i_1+i_2)-(r_1+r_2)$

$d=(i_1+i_2)-A$

$A+d=(i_1+i_2)$

$A+D=2i_1$ ; $D=$ Angle of minimum deviation.

$i_1=\frac{A+D}{2}$

$A=r_1+r_2$

$A=2r_1$ , as $r_1=r_2$

$r_1=\frac{A}{2}$

By snells law,

$\frac{\sin i_1}{\sin r_1}=\frac{n_2}{n_1}$

$\sin i_1=n.\sin r_1$ ; $n_1=$ Refractive index of air $=1$

$n=\frac{\sin i_1}{\sin r_1}$

$n=\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}$