A plano-convex lens has focal length 0.5 m and the refractive index of the material of the lens is 1.5. The radius of curvature of the convex surface of the lens is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.25 m
Given: focal length $f=0.5\ \mathrm{m}$, refractive index $\mu=1.5$.
Lens maker’s formula:
$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
For a plano-convex lens: one surface is plane $\Rightarrow R_2=\infty$, and convex surface radius = $R_1=R$.
So, $\frac{1}{f}=(\mu-1)\frac{1}{R}$
$R=(\mu-1)f$
Substitute values:
$R=(1.5-1)\times 0.5=0.5\times 0.5=0.25\ \mathrm{m}$
Final Answer: Radius of curvature $R=0.25\ \mathrm{m}=25\ \mathrm{cm}$