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JEE Main 202523 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Morning Shift
Subject: Physics
Chapter: Ray Optics

JEE Main 2025 · 23 January · Morning Shift · Physics · Ray Optics

A spherical surface of radius of curvature R , separates air from...

A spherical surface of radius of curvature

R

, separates air from glass (refractive index

=1.5

). The centre of curvature is in the glass medium. A point object '

O

' placed in air on the optic axis of the surface, so that its real image is formed at '

I

' inside glass. The line OI intersects the spherical surface at P and

\mathrm{PO}=\mathrm{PI}

. The distance PO equals to $A spherical surface of radius of curvature R , separates air from glass (refractive index =1.5 ). The centre of$

Options

A5 R
B3 R
C1.5 R
D2 R

Correct answer

A. 5 R

Step-by-step solution

$A spherical surface of radius of curvature R , separates air from glass (refractive index =1.5 ). The centre of$

\begin{aligned} & \mathrm{PO}=\mathrm{u}=-\mathrm{x} \\ & \mathrm{PI}=\mathrm{v}=\mathrm{x} \\ & \mathrm{PO}=\mathrm{PI} \\ & \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{\mathrm{x}}+\frac{1}{\mathrm{x}}=\frac{1}{2 \mathrm{R}} \\ & \frac{5}{2 \mathrm{x}}=\frac{1}{2 \mathrm{R}} \\ & \mathrm{X}=5 \mathrm{R}\end{aligned}

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