Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature 30 ~cm . The centre of curvature of surface is towards denser medium and a point object is placed on the principle axis in rarer medium at a distance of 15 ~cm from the pole of the surface. The distance of image from the pole of the surface is \underline\hspace2cm cm .

Geometrical Optics PYQ — JEE Main 2023 · 8 Apr · Shift 2 · Q59 with Solution

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Correct answer: 30

Given data

Refractive index of rarer medium: $n_1 = 1.0$
Refractive index of denser medium: $n_2 = 1.5$
Radius of curvature: $R = 30\,\text{cm}$
Object distance from pole: $15\,\text{cm}$

Since the object is in the rarer medium, it lies on the incident side.

Also, the centre of curvature is towards the denser medium, so with the usual Cartesian sign convention (positive to the right, direction of incident light taken left to right),

$u = -15\,\text{cm}$
$R = +30\,\text{cm}$

Formula for refraction at a spherical surface

The relation is

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

Substitute the values:

\frac{1.5}{v} - \frac{1.0}{-15} = \frac{1.5 - 1.0}{30}

\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30} = \frac{1}{60}

Solve for $v$

\frac{1.5}{v} = \frac{1}{60} - \frac{1}{15}

\frac{1.5}{v} = \frac{1 - 4}{60} = -\frac{3}{60} = -\frac{1}{20}

v = 1.5 \times (-20) = -30\,\text{cm}

Interpretation

The negative sign means the image is formed on the same side as the object (virtual image, in the rarer medium).

Hence, the distance of image from the pole is

|v| = 30\,\text{cm}

Final integer answer

\boxed{30}

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