A spherical interface of radius R separates t | vedclass

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(A) For refraction at a spherical surface, the formula is: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Given: $\mu_1 = 1$, $\mu_2 =

Vedclass · Accepted Answer

$1.66$

Vedclass · Answer

(A) For refraction at a spherical surface, the formula is: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Given: $\mu_1 = 1$, $\mu_2 = 1.4$, $u = -4R$, and the radius of curvature is $+R$ (as the center is in the second medium).Substituting the values: $\frac{1.4}{v} - \frac{1}{-4R} = \frac{1.4 - 1}{R} \implies \frac{1.4}{v} + \frac{1}{4R} = \frac{0.4}{R}$.$\frac{1.4}{v} = \frac{0.4}{R} - \frac{0.25}{R} = \frac{0.15}{R}$.$v = \frac{1.4R}{0.15} = \frac{140R}{15} = \frac{28R}{3} \approx 9.33R$.The lateral magnification $m$ for a spherical surface is given by $m = \frac{\mu_1 v}{\mu_2 u}$.$m = \frac{1 	imes (28R/3)}{1.4 	imes (-4R)} = \frac{28R/3}{-5.6R} = -\frac{28}{3 	imes 5.6} = -\frac{28}{16.8} = -1.666... \approx -1.67$.The magnitude of the magnification is $|m| = 1.67$.

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