JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Geometrical Optics

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

PO = PI

. The distance PO equals to-

(A)5 R✓ Correct
(B)3 R
(C)2 R
(D)1.5 R

Answer: (A)

Solution

Applying the refraction formula at a single spherical surface,

\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}

, with

μ_{1} = 1

(air) and

μ_{2} = 1.5

(glass).

Since

P O = P I

, take

u = - x

and

v = + x

\frac{1.5}{x} - \frac{1}{- x} = \frac{1.5 - 1}{R}

\frac{1.5}{x} + \frac{1}{x} = \frac{0.5}{R}

\frac{2.5}{x} = \frac{0.5}{R}

x = 5 R

P O = 5 R

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Geometrical Optics →

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Geometrical Optics

Solution

More JEE Main questions on Geometrical Optics