Question

A double convex lens made of glass (refractive index $n=1.5$) has both radii of curvature of magnitude $20cm$. Incident light rays parallel to the axis of the lens will converge at a distance $L$ such that

• A. L = 20 cm
• B. L = 10 cm
• C. L = 40 cm
• D. L = $\frac{20}{3}$ cm

Answer 1

Answer: (A)

Here, $n=1.5$ as per sign convention followed $R_1=+20cm$ and $R_2=-20cm$
$\therefore \frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$=(1.5-1)\left[\frac{1}{(+20)}-\frac{1}{(-20)}\right]$
$=0.5\times \frac{2}{20}=\frac{1}{20}$
$f=20cm$
Incident rays travelling parallel to the axis of lens will converge at its second principal focus. Hence,
$L=+20cm$