Q123 / 180 · KCET 2023 · Physics

An equiconvex lens made of glass of refractive index $\frac{3}{2}$ has focal length $f$ in air. It is completely immersed in water of refractive index $\frac{4}{3}$. The percentage change in the focal length is

  • A
    $400 \%$ increase
  • B
    $300 \%$ decrease
  • C
    $400 \%$ decrease
  • Correct answer
    D
    $300 \%$ increase

Solution

Given, ${ }^a \mu_g=\frac{3}{2}, f_{\text {air }}=f$ ${ }^a \mu_w=\frac{4}{3}$ Using lens Maker's formula, when lens is in air, $\begin{aligned} & \frac{1}{f_{\text {air }}}=\left({ }^a \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow \quad \frac{1}{f} & =\left(\frac{3}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow \quad \frac{1}{f} & =\frac{1}{2}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\end{aligned}$ When lens is immersed in water, then $\frac{1}{f_w}=\left({ }^w \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ Dividing Eq. (i) by Eq. (ii), we get $\begin{aligned} \quad \frac{f_w}{f} & =\frac{1}{2\left(\mu_g-1\right)} \\ \Rightarrow \quad f_w & =\frac{f}{2\left(\frac{\mu_g}{\mu_w}-1\right)}=\frac{f}{2\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)}=\frac{f}{\frac{1}{4}}=4 f \\ \Rightarrow \quad f_w & =4 f\end{aligned}$ $\therefore$ Percentage change in focal length $\begin{aligned} & =\frac{f_w-f}{f} \times 100 \\ & =\frac{4 f-f}{f} \times 100=300 \%\end{aligned}$

Hence, the correct answer is option: 'D'.

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