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 P O=P I . The distance P O equals to

Geometrical Optics PYQ — JEE Main 2025 · 23 Jan · Shift 1 · Q61 with Solution

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Correct answer: A

For refraction from medium 1 to medium 2 at a spherical surface,

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

Here:

Thus,

\frac{1.5}{v}-\frac{1}{u}=\frac{0.5}{R}

Since the line $OI$ meets the surface at $P$ , and object and image lie on the optic axis, $P$ is the pole. Given

PO=PI

So the object distance and image distance from the pole have equal magnitudes:

|u|=v

Because $u<0$ , this gives

v=-u

Put $u=-v$ into

\frac{1.5}{v}-\frac{1}{u}=\frac{0.5}{R}

Then

\frac{1.5}{v}-\frac{1}{-v}=\frac{0.5}{R}

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

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

Hence,

v=\frac{2.5}{0.5}R=5R

Since $PO=PI=v$ ,

PO=5R

The correct option is:

\boxed{5R}

So, Option A is correct.