Question

Focal length of a plano-convex lens is f cm. This is made up of a material of refractive index 2. If it is placed in the liquid of refractive index $\frac { 4 } { 3 }$ . Then its focal length will be

• A. f/2
• B. 2f
• C. f
• D. None

Answer 1

Answer: (B)

2f

Answer 2

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

\ = $\frac { 1 } { \mathrm { R } _ { 1 } }$ – $\frac { 1 } { \mathrm { R } _ { 2 } }$ … (I)

In the liquid;

$\frac { 1 } { \mathrm { f } ^ { \prime } }$ = $\left( \frac { 2 } { 4 / 3 } - 1 \right)$ $\left( \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right)$

$\frac { 1 } { \mathrm { f } ^ { \prime } }$ = 0.5 $\left( \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right)$ … (II)

Dividing I by II

= $\frac { \left( \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right) } { 0.5 \left( \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right) }$

$\frac { \mathrm { f } ^ { \prime } } { \mathrm { f } }$ = $\frac { 1 } { 0.5 }$ Ž f¢ = 2 f