Ray Optics

JEE Main level, single correct, medium

158 of 295

Question 158 of 295

For the thin convex lens, the radii of curvature are at 15 cm15 \mathrm{~cm} and 30 cm30 \mathrm{~cm} respectively. The focal length the lens is 20 cm20 \mathrm{~cm}. The refractive index of the material is :

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Answer: (C) 1.5

1f=(μlens μair −1)(1R1−1R2)⇒1+20=(μ1−1)(1+15−1(−30))⇒120=(μ−1)(330)⇒μ−1=12⇒μ=1+12=32=1⋅5\begin{aligned} & \frac{1}{\mathrm{f}}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {air }}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ & \Rightarrow \frac{1}{+20}=\left(\frac{\mu}{1}-1\right)\left(\frac{1}{+15}-\frac{1}{(-30)}\right) \\ & \Rightarrow \frac{1}{20}=(\mu-1)\left(\frac{3}{30}\right) \\ & \Rightarrow \mu-1=\frac{1}{2} \\ & \Rightarrow \mu=1+\frac{1}{2}=\frac{3}{2}=1 \cdot 5\end{aligned}

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