JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 10)
A biconvex lens of refractive index 1.5 has a focal length of $$20 \mathrm{~cm}$$ in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:
$$-$$16 cm
+16 cm
+160 cm
$$-$$160 cm
Explanation
$$\begin{aligned}
& \mu_1=1.5 \\
& \mu_m=1.6 \\
& f_a=20 \mathrm{~cm} \\
& \text { As } \frac{f_m}{f_a}=\frac{\left(\mu_1-1\right) \mu_m}{\left(\mu_1-\mu_m\right)} \\
& \frac{f_m}{20}=\frac{(1.5-1) 1.6}{(1.5-1.6)} \\
& f_m=-160 \mathrm{~cm}
\end{aligned}$$
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