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The refractive index of the material of a concave lens is $\mu$. It is immersed in a medium of refractive index $\mu_1$. A parallel beam of light is incident on the lens. The path of the emergent rays when $\mu_1>\mu$ is:
Options:
- A
- B
- C
- D
Solution:
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Verified Answer
The correct answer is:
If a lens of refractive index $\mu$ is immersed in a medium of refractive index $\mu_1$, then its focal length in medium is given by
$$
\frac{1}{\mathrm{f}_{\mathrm{m}}}=\left({ }_{\mathrm{m}} \mu_l-1\left(\frac{1}{\mathrm{R}_1}\right)-\frac{1}{\mathrm{R}_2}\right)
$$
If $\mathrm{f}_{\mathrm{a}}$ is the focal length of lens in air, then
$$
\begin{aligned}
&\frac{1}{\mathrm{f}_{\mathrm{a}}}=\left({ }_{\mathrm{a}} \mu_l-1\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\right. \\
&\Rightarrow \frac{\mathrm{f}_{\mathrm{m}}}{\mathrm{f}_{\mathrm{a}}}=\frac{\left({ }_{\mathrm{a}} \mu_l-1\right)}{\left({ }_{\mathrm{m}} \mu_l-1\right)}
\end{aligned}
$$
If $\mu_1>\mu$, then $\mathrm{f}_{\mathrm{m}}$ and $\mathrm{f}_{\mathrm{a}}$ have opposite signs and the nature of lens changes i.e. a convex lens diverges the light rays and concave lens converges the light rays. Thus given option (a) is correct.
$$
\frac{1}{\mathrm{f}_{\mathrm{m}}}=\left({ }_{\mathrm{m}} \mu_l-1\left(\frac{1}{\mathrm{R}_1}\right)-\frac{1}{\mathrm{R}_2}\right)
$$
If $\mathrm{f}_{\mathrm{a}}$ is the focal length of lens in air, then
$$
\begin{aligned}
&\frac{1}{\mathrm{f}_{\mathrm{a}}}=\left({ }_{\mathrm{a}} \mu_l-1\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\right. \\
&\Rightarrow \frac{\mathrm{f}_{\mathrm{m}}}{\mathrm{f}_{\mathrm{a}}}=\frac{\left({ }_{\mathrm{a}} \mu_l-1\right)}{\left({ }_{\mathrm{m}} \mu_l-1\right)}
\end{aligned}
$$
If $\mu_1>\mu$, then $\mathrm{f}_{\mathrm{m}}$ and $\mathrm{f}_{\mathrm{a}}$ have opposite signs and the nature of lens changes i.e. a convex lens diverges the light rays and concave lens converges the light rays. Thus given option (a) is correct.
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