Question

The radii of curvature of the surface of double convex lens are 20 cm and 40 cm respectively, and its focal length is 20 cm. What is the refractive index of the material of the lens?

• A. $\frac { 5 } { 2 }$
• B. $\frac { 4 } { 3 }$
• C. $\frac { 5 } { 3 }$
• D. $\frac { 4 } { 5 }$

Answer 1

Answer: (C)

$\frac { 5 } { 3 }$

Answer 2

: $\frac { 1 } { \mathrm { f } } = ( \mu - 1 ) \left( \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right)$

Here, $\mathrm { R } _ { 1 } = 20 \mathrm {~cm} , \mathrm { R } _ { 2 } = - 40 \mathrm {~cm} , \mathrm { f } = 20 \mathrm {~cm}$

$\therefore \frac { 1 } { 20 } = ( \mu - 1 ) \left( \frac { 1 } { 20 } + \frac { 1 } { 40 } \right)$

$\frac { 1 } { 20 } = ( \mu - 1 ) \frac { 3 } { 40 } \Rightarrow ( \mu - 1 ) = \frac { 2 } { 3 }$ or $\mu = \frac { 2 } { 3 } + 1$

$\mu = \frac { 5 } { 3 }$