Question

The radius of curvature of each surface of a convex lens of refractive index $1.5$ is $40cm$. Its power is

• A. $2.5D$
• B. $2D$
• C. $1.5D$
• D. $1D$

Answer 1

Answer: (A)

Power, $P=\frac{1}{f}=(\mu -1)[\frac{1}{R_1}-\frac{1}{R_2}]$
Here, $\mu =1.5,R_1=40cm=0.4cm$,
$R_2=-0.4m$
$P=(1.5-1)(\frac{1}{0.4}+\frac{1}{0.4})$
$=2.5D$