Question

If the radius of curvature of the curved surface of a plano-convex lens is $50 \, cm$ , its focal length is $\left(\mu = 1.5\right)$

• A. $0.5 \, m$
• B. $0.75 \, m$
• C. $0.25 \, m$
• D. $1 \, m$

Answer 1

Answer: (D)

From lens maker's formula,
$\frac{1}{f}=\left(n - 1\right) \, \left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)$
Given, $n=1.5$
$R_{1}=\infty \,$ (Plane surface)
$R_{2}= \, -50 \, cm=-0.5 \, m$
$\frac{1}{f}=\left(1.5 - 1\right)\times \frac{1}{0.5}$
$=0.5\times \frac{1}{0.5}=1 \, m$