Question

A thin lens made of glass (refractive index $1.5$) of focal length $f=16cm$ is immersed in a liquid of refractive index $1.42$. If its focal length in liquid is $f_l$, then the ratio $\frac{f_l}{f}$ is closest to the integer.

• A. $9$
• B. $17$
• C. $1$
• D. $5$

Answer 1

The correct option is A

$9$

Step1: Given data and assumptions.

Refractive index of the glass, ${\mu }_g=1.5$

Refractive index of medium, ${\mu }_l=1.42$

The focal length of the lens, $f=16cm$

The focal length in the liquid $=f_l$

Also, we know the refractive index of the air, ${\mu }_a=1$

Step2: Formula used:

$\frac{1}{f}=\left(\frac{{\mu }_2}{{\mu }_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Where, $f$ is the focal length and $R_1$ and $R_2$ are the radius of curvature.

Step3: Find the ratio of the focal length in the liquid to the focal length of the lens.

Now,

$\frac{1}{f}=\left(\frac{{\mu }_g}{{\mu }_a}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Where $f$ is the focal length of the lens in air.

$\Rightarrow$$\frac{1}{f}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\Rightarrow$$\frac{1}{f}=0.5\times \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ……$\left(i\right)$

And,

$\Rightarrow$$\frac{1}{f_l}=\left(\frac{{\mu }_g}{{\mu }_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\Rightarrow$$\frac{1}{f_l}=\left(\frac{1.5}{1.42}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\Rightarrow$$\frac{1}{f_l}=0.056\times \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ……$\left(ii\right)$

Where $f_l$ is the focal length of the lens in the liquid.

By dividing the equation $\left(ii\right)$ by $\left(i\right)$ we get,

$\Rightarrow$$\frac{f}{f_l}=\frac{0.056}{0.5}$

$\Rightarrow$$\frac{f_l}{f}=\frac{0.5}{0.056}$

$\Rightarrow$$\frac{f_l}{f}=8.928\approx 9$

Thus, the ratio $\frac{{\mathbf{f}}_{\mathbf{l}}}{\mathbf{f}}\mathbf{=}\mathbf{9}$

Hence, option A is correct.