Question

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

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

Answer 1

Answer: (A)

2.5 D

Answer 2

: Power, $\mathrm { P } = \frac { 1 } { \mathrm { f } } = ( \mu - 1 ) \left[ \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right]$

Here, $\mu = 1.5 , \mathrm { R } _ { 1 } = 40 \mathrm {~cm} = 0.4 \mathrm {~cm}$

$= ( 1.5 - 1 ) \left( \frac { 1 } { 0.4 } + \frac { 1 } { 0.4 } \right)$

$\mathrm { P } = 2.5 \mathrm { D }$