A spherical surface of radius of curvature R, separates air from glass (refreactive 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 '' ins

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Solution

Q.
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 O $I$ intersects the spherical surface at P and PO = P $I$ . The distance PO equals to: [2025]

1 5 R

2 3 R

3 2 R

4 1.5 R

Ans.
(1)

PO = u = –x

P $I$ = v = x

PO = P $I$

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

$\frac{1.5}{x} + \frac{1}{x} = \frac{1}{2 R}$

$\frac{5}{2 x} = \frac{1}{2 R} \Rightarrow x = 5 R$

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