For the thin convex lens, the radii of curvature are at 15 ~cm and 30 ~cm respectively. The focal length the lens is 20 ~cm . The refractive index of the material is :

Correct answer: C. 1.5 aligned & 1 f =( μ_ lens μ_ air -1)( 1 R_1 - 1 R_2 ) & (1)/(+20)=((μ)/(1)-1)((1)/(+15)-(1)/((-30))) & (1)/(20)=(μ-1)((3)/(30)) & μ-1=(1)/(2) & μ=1+(1)/(2)=(3)/(2)=1 · 5 aligned

JEE Main 20246 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Evening Shift
Subject: Physics
Chapter: Ray Optics

JEE Main 2024 · 6 April · Evening Shift · Physics · Ray Optics

For the thin convex lens, the radii of curvature are at 15 ~cm and...

For the thin convex lens, the radii of curvature are at

15 \mathrm{~cm}

and

30 \mathrm{~cm}

respectively. The focal length the lens is

20 \mathrm{~cm}

. The refractive index of the material is :

Options

A1.2
B1.8
C1.5
D1.4

Correct answer

C. 1.5

Step-by-step solution

\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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