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…

Question

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 $\mathrm{PO}=\mathrm{PI}$. The distance PO equals to

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{PO}=\mathrm{u}=-\mathrm{x} \\ & \mathrm{PI}=\mathrm{v}=\mathrm{x} \\ & \mathrm{PO}=\mathrm{PI} \\ & \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{\mathrm{x}}+\frac{1}{\mathrm{x}}=\frac{1}{2 \mathrm{R}} \\ & \frac{5}{2 \mathrm{x}}=\frac{1}{2 \mathrm{R}} \\ & \mathrm{X}=5 \mathrm{R}\end{aligned}$

A spherical surface of radius of curvature $R$, separates air from glass (refractive index $=1.5$). The…

Solution