Question 216: Ray Optics

Question

A convex lens of refractive index 1.5 and focal length $f=18 \mathrm{~cm}$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha 	imes \mathrm{f}$. The value of $\alpha$ is $\_\_\_\_$ .
(refractive index of water $=4 / 3$ )

Accepted Answer

The Lens Maker's Formula is
$$ \frac{1}{\mathrm{f}}=\left(\frac{\mu_1}{\mu_{\mathrm{m}}}-1ight)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}ight) $$
where $\mu_1$ is the refractive index of the lens and $\mu_m$ is the refractive index of the surrounding medium.
Case-1 Focal length in Air $\left(\mathrm{f}_{\mathrm{a}}ight)$
In air, $\mu_{\mathrm{m}}=1$.
$$ \frac{1}{\mathrm{f}_{\mathrm{a}}}=\left(\frac{\mu_1}{1}-1ight)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}ight) $$
Let $\mathrm{K}=\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}ight) \Rightarrow \frac{1}{\mathrm{f}_{\mathrm{a}}}=\left(\mu_1-1ight) \mathrm{K}$
Given $\mu_1=1.5$ and $\mathrm{f}_{\mathrm{a}}=\mathrm{f}=18 \mathrm{~cm}$ :
$$ \frac{1}{\mathrm{f}}=(1.5-1) \mathrm{K}=0.5 \mathrm{~K} \Rightarrow \mathrm{~K}=\frac{2}{\mathrm{f}} $$
Case-2 Focal length in Water ( $\mathrm{f}_{\mathrm{w}}$ )
In water, $\mu_{\mathrm{m}}=\mu_{\mathrm{w}}=4 / 3$ :
$\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{\mu_{\mathrm{l}}}{\mu_{\mathrm{w}}}-1ight) \mathrm{K}$
$$\Rightarrow $$ $\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{1.5}{4 / 3}-1ight) \mathrm{K}=\left(\frac{4.5}{4}-1ight) \mathrm{K}=\left(\frac{9}{8}-1ight) \mathrm{K}$
$$ \frac{1}{\mathrm{f}_{\mathrm{w}}}=\frac{1}{8} \mathrm{~K}=\frac{1}{8} 	imes(2 \mathrm{f})=\frac{1}{4} \mathrm{f} $$
Therefore, $\mathrm{f}_{\mathrm{w}}=4 \mathrm{f}$.
The difference in focal lengths is given as :
$$ \Delta \mathrm{f}=\mathrm{f}_{\mathrm{w}}-\mathrm{f}_{\mathrm{a}} $$
$$\Rightarrow $$ $$ \Delta \mathrm{f}=4 \mathrm{f}-\mathrm{f}=3 \mathrm{f}=\alpha 	imes \mathrm{f} \Rightarrow \alpha=3 $$
Therefore, the value of $\alpha$ is 3.
Hence, the correct answer is 3.

Ray Optics

A convex lens of refractive index 1.5 and focal length f=18 \mathrm{~cm} is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is \alpha \times \mathrm{f}. The value of \alpha is \_\_\_\_ . (refractive index of water =4 / 3 )

Integer / Numeric Answer Type