A point object kept at P in front of a transparent sphere of radius 'R' has its image formed by the first refracting surface at Q such that PO = QO. The refractive index of the material of the sphere is 3. The distance PO is equal to

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For refraction at a single spherical surface, the formula is given by:

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

Applying the Cartesian sign convention:
Object distance, $u = - PO$
Image distance, $v = + OQ = + PO$ (since it is given that $PO = QO$ )
Radius of curvature of the first surface, $R = + R$
Refractive index of the surrounding medium (air), $μ_{1} = 1$
Refractive index of the sphere, $μ_{2} = 3$

Substituting these values into the formula, we get:

$\frac{3}{PO} - \frac{1}{- PO} = \frac{3 - 1}{R}$

$\frac{3}{PO} + \frac{1}{PO} = \frac{2}{R}$

$\frac{4}{PO} = \frac{2}{R}$

$PO = 2 R$

Answer: 2R

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