The first focal length f1 for refraction at a spherical surface is defined as the value of u corresponding to v=∞(as shown) with refractive indices of two mediums, as n1 and n2 The second focal length f2 is defined as value of v for u=∞

n2v−n1u=n2−n1Rn2f2−n1∞=n2−n1Rf2=n2Rn2−n1Similarly n2∞−n1f1=n2−n1Rf1=−n1Rn2−n1

The first focal length f1 for refraction at a spherical surface is defined as the value of u corresponding to v=∞(as shown) with refractive indices of two mediums, as n1 and n2 The second focal length f2 is defined as value of v for u=∞

n2v−n1u=n2−n1Rn2f2−n1∞=n2−n1Rf2=n2Rn2−n1Similarly n2∞−n1f1=n2−n1Rf1=−n1Rn2−n1

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The first focal length $f_{1}$ for refraction at a spherical surface is defined as the value of $u$ corresponding to $v = \infty (a s s h o w n)$ with refractive indices of two mediums, as $n_{1}$ and $n_{2}$ The second focal length $f_{2}$ is defined as value of v for $u = \infty$

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$f_{1}$ is equal to $- \frac{n_{1} R}{(n_{2} - n_{1})}$

$f_{2}$ is equal to $- \frac{n_{1} R}{(n_{2} - n_{1})}$

$f_{2}$ is equal to $\frac{n_{2} R}{(n_{2} - n_{1})}$

$f_{1}$ is equal to $\frac{n_{2} R}{(n_{2} - n_{1})}$

answer is A, D.

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Detailed Solution

$\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}$

$\begin{array}{l} \frac{n_{2}}{f_{2}} - \frac{n_{1}}{\infty} = \frac{n_{2} - n_{1}}{R} \\ f_{2} = \frac{n_{2} R}{n_{2} - n_{1}} \end{array}$

Similarly $\frac{n_{2}}{\infty} - \frac{n_{1}}{f_{1}} = \frac{n_{2} - n_{1}}{R}$

$f_{1} = - \frac{n_{1} R}{n_{2} - n_{1}}$

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The first focal length f1 for refraction at a spherical surface is defined as the value of u corresponding to v=∞(as shown) with refractive indices of two mediums, as n1 and n2 The second focal length f2 is defined as value of v for u=∞

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The first focal length $f_{1}$ for refraction at a spherical surface is defined as the value of $u$ corresponding to $v = \infty (a s s h o w n)$ with refractive indices of two mediums, as $n_{1}$ and $n_{2}$ The second focal length $f_{2}$ is defined as value of v for $u = \infty$