Question

If the radius of curvature of the curved surface of a planoconvex lens is $50cm$, its focal length is $(\mu =1.5)$

• A. $-0.5m$
• B. $-0.75m$
• C. $-1.25m$
• D. $-0.25m$
• E. $1m$

Answer 1

The correct option is E $1m$

From lens formula,
$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Given, $n=15$
$R_1=\infty$ (Plane surface)
$R_2=-50cm=-0.5m$
$\frac{1}{f}=(1.5-1)\times \frac{1}{0.5}$
$=0.5\times \frac{1}{0.5}=1$m.