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

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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.

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