JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Geometrical Optics

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A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure.
Distance of the image of an object 'O', is:
(C is the center of curvature of the spherical surface and

R

is the radius of curvature)

(A)0.24 m right to the spherical surface
(B)0.4 m left to the spherical surface✓ Correct
(C)0.24 m left to the spherical surface
(D)0.4 m right to the spherical surface

Answer: (B)

Solution

Refraction at the single surface obeys

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

. With

μ_{1} = 1

μ_{2} = 1.5

, object distance

u = - 0.2

m, and the centre of curvature lying on the transmission side so

R = + 0.4

\frac{1.5}{v} + 5 = \frac{0.5}{0.4} = 1.25

, which gives

\frac{1.5}{v} = - 3.75

and

v = - 0.4

m. The image forms 0.4 m to the left of the surface, on the same side as the object.

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Geometrical Optics

Solution

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