Ray Optics

JEE Main level, single correct, medium

137 of 295

Question 137 of 295

A lens having refractive index 1.6 has focal length of 12 cm , when it is in air. Find the focal length of the lens when it is placed in water.
(Take refractive index of water as 1.28)

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Answer: (B) 288 mm

As we know,
1f=[μLμm−1][1R1−1R2]\frac{1}{\mathrm{f}}=\left[\frac{\mu_{\mathrm{L}}}{\mu_{\mathrm{m}}}-1\right]\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]
For air μm=1\mu_{\mathrm{m}}=1
112=[1.6−1][1R1−1R2]112=610[1R1−1R2][1R1−1R2]=1072\begin{aligned} & \frac{1}{12}=[1.6-1]\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\ & \frac{1}{12}=\frac{6}{10}\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\ & {\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]=\frac{10}{72}}\end{aligned}
For water
1f=[1.61.28−1][1072]=32128×1072\frac{1}{\mathrm{f}}=\left[\frac{1.6}{1.28}-1\right]\left[\frac{10}{72}\right]=\frac{32}{128} \times \frac{10}{72}
1f=14×1072f=28.8 cmf=288 mm\begin{aligned} & \frac{1}{\mathrm{f}}=\frac{1}{4} \times \frac{10}{72} \\ & \mathrm{f}=28.8 \mathrm{~cm} \\ & \mathrm{f}=288 \mathrm{~mm}\end{aligned}

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