Q142 / 160 · AP EAMCET 2024 (22 May Shift 2) · Physics
When a convex lens is immersed in a liquid of refractive index equal to $80 \%$ of the refractive index of the material of the lens, the focal length of the lens increases by $100 \%$. The refractive index of the liquid is
- A1.27
- Correct answerB1.2
- C1.33
- D1.4
Solution
When convex lens placed in air,
$\frac{1}{\mathrm{f}_{\mathrm{a}}}=\left(\mu_1-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$ ...(i)
When convex lens immersed in medium, $\frac{1}{\mathrm{f}_{\mathrm{m}}}=\left(\frac{\mu_1}{\mu_{\mathrm{m}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$ ...(ii)
Dividing eq(i) by eq(ii), we get $\begin{aligned} & \frac{f_m}{f_a}=\frac{\left(\mu_1-1\right)}{\left(\frac{\mu_1}{\mu_m}-1\right)}=\frac{\left(\frac{5}{4} \mu_m-1\right)}{\left(\frac{5}{4}-1\right)} \\ & \Rightarrow \frac{2 f_a}{f_a}=4\left(\frac{5}{4} \mu_m-1\right)=5 \mu_m-4 \\ & \therefore \mu_m=1.2 \end{aligned}$
When convex lens immersed in medium, $\frac{1}{\mathrm{f}_{\mathrm{m}}}=\left(\frac{\mu_1}{\mu_{\mathrm{m}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$ ...(ii)
Dividing eq(i) by eq(ii), we get $\begin{aligned} & \frac{f_m}{f_a}=\frac{\left(\mu_1-1\right)}{\left(\frac{\mu_1}{\mu_m}-1\right)}=\frac{\left(\frac{5}{4} \mu_m-1\right)}{\left(\frac{5}{4}-1\right)} \\ & \Rightarrow \frac{2 f_a}{f_a}=4\left(\frac{5}{4} \mu_m-1\right)=5 \mu_m-4 \\ & \therefore \mu_m=1.2 \end{aligned}$
Hence, the correct answer is option: 'B'.