Light from a point source in air falls on a spherical glass surface (refractive index, \mu=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 \underline\hspace2cm m.

Geometrical Optics PYQ — JEE Main 2025 · 3 Apr · Shift 2 · Q72 with Solution

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Correct answer: 4

Use refraction at a spherical surface formula

For refraction from medium $n_1$ to medium $n_2$ at a spherical surface,

\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}

Here:

object is in air: $n_1=1$
image is in glass: $n_2=1.5$
image distance inside glass: $v=+200\text{ cm}$
radius of curvature: $R=+50\text{ cm}$

We take distances positive in the direction of incident light (towards glass). Since the image is formed inside glass, $v$ is positive. For this convex surface as seen from air, $R$ is also positive.

Substitute the values

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

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

\frac{3}{400}-\frac{1}{u}=\frac{1}{100}

Solve for $u$

-\frac{1}{u}=\frac{1}{100}-\frac{3}{400}

-\frac{1}{u}=\frac{4-3}{400}=\frac{1}{400}

u=-400\text{ cm}

So the object is $400$ cm from the surface on the air side.

Convert to metres

400\text{ cm}=4\text{ m}

Therefore, the magnitude of the distance of the source from the glass surface is:

\boxed{4}

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