Ray Optics

JEE Main level, numerical answer, medium

128 of 295

Question 128 of 295

A convex lens of refractive index 1.5 and focal length f=18 cmf=18 \mathrm{~cm} is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is α×f\alpha \times \mathrm{f}. The value of α\alpha is ____\_\_\_\_.
(refractive index of water =4/3=4 / 3)

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Answer: 3

Using the lensmaker's equation, in air:
1fair=(n−1)(1R1−1R2)\frac{1}{f_{air}} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) where (n−1)=0.5(n-1) = 0.5 for glass with n=1.5n = 1.5.
In water:
1fwater=(nnwater−1)(1R1−1R2)\frac{1}{f_{water}} = \left(\frac{n}{n_{water}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)
=(1.54/3−1)(1R1−1R2)= \left(\frac{1.5}{4/3} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)
=18(1R1−1R2)= \frac{1}{8}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)
Since (1R1−1R2)=2fair=218\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{2}{f_{air}} = \frac{2}{18},
We get 1fwater=18⋅218=172\frac{1}{f_{water}} = \frac{1}{8} \cdot \frac{2}{18} = \frac{1}{72}, so fwater=72f_{water} = 72 cm.
The difference is fwater−fair=72−18=54f_{water} - f_{air} = 72 - 18 = 54 cm.
Therefore α=5418=3\alpha = \frac{54}{18} = 3.

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