When a convex lens is immersed in two different liquids of refractive indices 1.25 and 1.5. The ratio of the…

When a convex lens is immersed in two different liquids of refractive indices 1.25 and 1.5. The ratio of the focal lengths of the lens is $5: 16$. The refractive index of the material of the lens is
  1. 1.55
  2. 1.5
  3. 1.65
  4. 1.6

Solution

In first liquid, $\mu_1=1.25$ $\therefore \frac{1}{\mathrm{f}_1}=\left(\frac{\mu_2}{1.25}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$
In second liquid, $\mu_1=1.5$ $\begin{aligned} & \therefore \frac{1}{\mathrm{f}_2}=\left(\frac{\mu_2}{1.5}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ & \therefore \frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{\left(\frac{\mu_2}{1.5}-1\right)}{\left(\frac{\mu_2}{1.25}-1\right)}=\frac{5}{16} \\ & \therefore \quad \mu_2=1.65 \end{aligned}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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