Ray Optics

JEE Main level, single correct, medium

141 of 295

Question 141 of 295


A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object ' O ', is :
( C is the center of curvature of the spherical surface and RR is the radius of curvature)

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Answer: (B) 0.4 m left to the spherical surface

μ2v−μ1u=μ2−μ1R1.5v−1(−0.2)=1.5−10.41.5v=0.50.4−10.21.5v=−1.50.4v=−0.4 m\begin{aligned} & \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{\mathrm{v}}-\frac{1}{(-0.2)}=\frac{1.5-1}{0.4} \\ & \frac{1.5}{\mathrm{v}}=\frac{0.5}{0.4}-\frac{1}{0.2} \\ & \frac{1.5}{\mathrm{v}}=-\frac{1.5}{0.4} \\ & \mathrm{v}=-0.4 \mathrm{~m}\end{aligned}

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