Light from a point source in air falls on a spherical glass surface (refractive index, \(\mu=1.5\) and radius of curvature \(=50\,\text{cm}\)). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is \_\_\_\_ m.

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Geometrical Optics

Light from a point source in air falls on a spherical glass surface (refractive index,

μ = 1.5

and radius of curvature

= 50 cm

). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ____ m.

Answer: 4

Solution

Refraction at a single spherical surface relates object and image distances through

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

. Light travels from air into glass, so

μ_{1} = 1

and

μ_{2} = 1.5

, with the image inside the glass at

v = + 200 cm

and the convex surface facing the source giving

R = + 50 cm

. Substituting,

\frac{1.5}{200} - \frac{1}{u} = \frac{0.5}{50} = \frac{1}{100}

, so

\frac{1}{u} = \frac{1.5}{200} - \frac{1}{100} = - \frac{1}{400}

, which gives

u = - 400 cm

. The source lies

4 m

from the glass surface.

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Integer answer

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JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Geometrical Optics

Solution

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