A deviation of 2^{circ} is produced in the ye | vedclass

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(D) Given: $\omega_{1} = 0.02, \mu_{1} = 1.5$ (Crown glass) and $\omega_{2} = 0.03, \mu_{2} = 1.6$ (Flint glass).For an achromatic combination,the net

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$12$

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(D) Given: $\omega_{1} = 0.02, \mu_{1} = 1.5$ (Crown glass) and $\omega_{2} = 0.03, \mu_{2} = 1.6$ (Flint glass).For an achromatic combination,the net dispersion is zero,so $	heta_{1} = 	heta_{2}$.This implies $\omega_{1} \delta_{1} = \omega_{2} \delta_{2}$,where $\delta$ is the deviation.The net deviation is given as $\delta_{	ext{net}} = \delta_{1} - \delta_{2} = 2^{\circ}$.From the dispersion equation,$\delta_{2} = \frac{\omega_{1}}{\omega_{2}} \delta_{1} = \frac{0.02}{0.03} \delta_{1} = \frac{2}{3} \delta_{1}$.Substituting this into the net deviation equation: $\delta_{1} - \frac{2}{3} \delta_{1} = 2^{\circ}$.$\frac{1}{3} \delta_{1} = 2^{\circ} \implies \delta_{1} = 6^{\circ}$.The deviation produced by a prism is $\delta = (\mu - 1)A$.For the crown glass prism: $6^{\circ} = (1.5 - 1) A_{1}$.$6^{\circ} = 0.5 A_{1} \implies A_{1} = 12^{\circ}$.

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