Question

An equiconvex lens has power $P$ and radii of curvature $R$. The radius of curvature of the surface of a plano-convex lens made of the same material with power $1.5P$ is

• A. $2R$
• B. $3R$
• C. $\frac{R}{3}$
• D. $\frac{R}{2}$

Answer 1

The correct option is C $\frac{R}{3}$

Given that,
Radius of curvature of double convex lens $=R$
Power of lens $=P$
Assuming that the refractive index of lens is ${\mu }_l$
By using the formula
$\frac{1}{f}=({\mu }_l-1)(\frac{1}{R}-(-\frac{1}{R}))$
$P=({\mu }_l-1)\left(\frac{2}{R}\right)\dots \dots \left(1\right)$

For plano-convex lens,
$R_1=\infty ,R_2=-R^{\prime }$ (Assume)
The required power of plano-convex lens $=1.5P$
So,
$1.5P=({\mu }_l-1)(\frac{1}{\infty }-\frac{1}{-R^{\prime }})$
$\frac{3}{2}P=({\mu }_l-1)\left(\frac{1}{R^{\prime }}\right)\dots \dots \left(2\right)$
Divide $\left(1\right)$ by $\left(2\right)$
$\frac{P}{\frac{3}{2}P}=\frac{\frac{2}{R}}{\frac{1}{R^{\prime }}}$
$R^{\prime }=\frac{R}{3}$