Ray Optics

JEE Main level, numerical answer, medium

168 of 295

Question 168 of 295

The radius of curvature of each surface of a convex lens having refractive index 1.81.8 is 20cm20cm. The lens is now immersed in a liquid of refractive index 1.51.5. The ratio of power of lens in air to its power in the liquid will be x:1x:1. The value of xx is

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Answer: 4

The power of a lens is given by
P1=1f{P}_{1}=\dfrac{1}{f}
Using Lens Maker's formula,
P1=(1.8−1)(1R1−1R2){P}_{1}=(1.8-1)\left(\dfrac{1}{{R}_{1}}-\dfrac{1}{{R}_{2}}\right) and
P2=1fimmersed=(1.81.5−1)(1R1−1R2){P}_{2}=\dfrac{1}{{f}_{\text{immersed}}}=\left(\dfrac{1.8}{1.5}-1\right)\left(\dfrac{1}{{R}_{1}}-\dfrac{1}{{R}_{2}}\right)
Hence, the ratio is,
P1P2=(1.8−1)(1.81.5−1)=0.8×1.50.3=4\dfrac{{P}_{1}}{{P}_{2}}=\dfrac{(1.8-1)}{\left(\dfrac{1.8}{1.5}-1\right)}=\dfrac{0.8\times 1.5}{0.3}=4

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