# The dispersive power of crown and flint glasses are 0.03 and 0.05 respectively . If difference in refractive indices of blue and red colours is 0.014 for crown glass and 0.023 for flint glass . Then the angle of two prisms for a deviation of $10^{\circ}$ (with out dispersion )

$(a)\;26.8^{\circ}\; and\; 54.9^{\circ}\\ (b)\;51^{\circ}\; and\; 26^{\circ} \\ (c)\;53.6 ^{\circ}\; and\; 32.6^{\circ} \\ (d)\;None$

The crown glass
$w=0.03$ and $y_b-y_r =0.0K$
We know that, $w= \large\frac{\mu_b -\mu r}{\mu-1}$
$\therefore (\mu-1)=\large\frac{0.014}{0.03}$$=0.4667 For flint glass, w'= 0.05 \mu _b'- \mu_r=0.023 \therefore \mu'-1=\large\frac{0.023}{0.05}$$=0.46$
Again $\large\frac{\delta }{\delta '}=\frac{\omega '}{\omega}$
$\qquad= \large\frac{0.05}{0.03}=\frac{5}{3}$
$\delta =\large\frac{5}{3}$$\delta '$-----(1)
Given that , $\delta - \delta'=10$
$\large\frac{5}{3} \delta '- \delta' =10$-----(2)
From (1) and (2)
$\delta'= 15^{\circ}\;and \; \delta =25^{\circ}$
Now using formula $\delta = (\mu-1) A$
A (for crown glass) = $\large\frac{\delta}{(\mu-1)}=\frac{25}{0.4667}$
$\qquad =53.6 ^{\circ}$
A' (for flint glass) = $\large\frac{\delta'}{(\mu' -1)}=\frac{15}{0.46}$
$\qquad= 32.6^{\circ}$
Hence c is the correct answer.