Question

A spherical surface of radius of curvature $R$ separates air (refractive index $1.0$) from glass (refractive index $1.5$). The center of curvature is in the glass. A point object $P$ placed in air is found to have a real image $Q$ in the glass. The line $PQ$ cuts the surface at a point $O$, and $PO=OQ$. The distance $PQ$ is equal to

• A. $5R$
• B. $3R$
• C. $2R$
• D. $1.5R$

Answer 1

Answer: (A)

$5R$

Let the object distance be $x$ i.e. $u=-x$

As $PO=OQ$, thus image distance is also equal to $x$ i.e. $v=x$

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

$\therefore$ $\frac{1.5}{x}-\frac{1}{-x}=\frac{1.5-1}{R}$

Or, $\frac{2.5}{x}=\frac{0.5}{R}$ $⟹x=PO=5R$