JEE MAIN - Physics (2025 - 8th April Evening Shift - No. 3)
A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be
$ \frac{700}{11} $ cm
$ \frac{600}{11} $ cm
$ \frac{800}{11} $ cm
$ \frac{500}{11} $ cm
Explanation
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$$\begin{aligned} \frac{1}{\mathrm{f}} & =\left(\frac{1.3-1}{1}\right)\left(\frac{1}{\infty}-\frac{1}{-30}\right) \\ & =\left(\frac{1.5-1}{1}\right)\left(\frac{1}{-30}-\frac{1}{-30}\right) \\ & =\frac{0.3}{30}+\frac{0.5}{60}=\frac{1}{100}+\frac{1}{120} \\ & =\frac{6+5}{600}=\frac{11}{600} \\ \mathrm{f} & =\frac{600}{11} \mathrm{~cm} \end{aligned}$$
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