JEE Main 2025 Jan 23 Shift 1, Physics Q5: Refraction at Plane and Spherical surfaces

JEE Main2025Jan 23, Shift 1Physics

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
$5 R$
B
$3 R$
C
$2 R$
D
$1.5 R$

Solution

Use the refraction-at-spherical-surface formula with light travelling from air ( $μ_{1} = 1$ ) into glass ( $μ_{2} = 1.5$ ). Since the centre of curvature lies in the glass (the denser medium), $R$ is positive in the standard sign convention.

Take P as origin, light going left-to-right. With PO = PI = $x$ : $u = - x$ , $v = + x$ .

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

$\frac{1.5}{x} - \frac{1}{- x} = \frac{0.5}{R} \Rightarrow \frac{2.5}{x} = \frac{0.5}{R} \Rightarrow x = 5 R$ .

$∴$ PO = $5 R$ .

Concept behind this question

Refraction at Plane and Spherical surfacesNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 23 Shift 1, Physics Q5: Refraction at Plane and Spherical surfaces

Concept behind this question

More previous year questions on Refraction at Plane and Spherical surfaces

Practice more Physics