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(D) For refraction at a spherical surface,the formula is: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.For the concave surface in fig

Vedclass · Accepted Answer

Both $(A)$ and $(B)$.

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(D) For refraction at a spherical surface,the formula is: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.For the concave surface in figure $B$,the pole is at the origin,light travels from left to right,so $u = -x$ and $R = -R$ (sign convention).The equation becomes: $\frac{\mu_2}{v} - \frac{\mu_1}{-x} = \frac{\mu_2 - \mu_1}{-R} \implies \frac{\mu_2}{v} = \frac{\mu_1 - \mu_2}{R} - \frac{\mu_1}{x}$.For a virtual image,$v Case $1$: If $\mu_2 > \mu_1$,then $\frac{\mu_2}{v} = -\frac{(\mu_2 - \mu_1)}{R} - \frac{\mu_1}{x}$. Since both terms are negative,$v$ is always negative,so a virtual image is formed for any $x$.Case $2$: If $\mu_2  0$. Then $\frac{\mu_2}{v} = \frac{\Delta\mu}{R} - \frac{\mu_1}{x}$. For a virtual image,$v Thus,both statements $(A)$ and $(B)$ are correct.

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