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Question: Shown in figure here is a convergent lens placed inside a cell filled with a liquid. The lens has fo...

Shown in figure here is a convergent lens placed inside a cell filled with a liquid. The lens has focal length +20 cm when in air and its material has refractive index 1.50. If the liquid has refractive index 1.60, the focal length of the system is

A
  • 80 cm
B

– 80 cm

C

– 24 cm

D

– 100 cm

Answer

– 100 cm

Explanation

Solution

Here 1f1=(1.6−1)(1∞−120)=−3100\frac { 1 } { f _ { 1 } } = ( 1.6 - 1 ) \left( \frac { 1 } { \infty } - \frac { 1 } { 20 } \right) = \frac { - 3 } { 100 } .......(i)

1f2=(1.5−1)(120−1−20)=120\frac { 1 } { f _ { 2 } } = ( 1.5 - 1 ) \left( \frac { 1 } { 20 } - \frac { 1 } { - 20 } \right) = \frac { 1 } { 20 } .......(ii)

1f3=(1.6−1)(1−20−1∞)=−3100\frac { 1 } { f _ { 3 } } = ( 1.6 - 1 ) \left( \frac { 1 } { - 20 } - \frac { 1 } { \infty } \right) = \frac { - 3 } { 100 } .......(iii)

By using 1F=1f1+1f2+1f3\frac { 1 } { F } = \frac { 1 } { f _ { 1 } } + \frac { 1 } { f _ { 2 } } + \frac { 1 } { f _ { 3 } } ⇒ 1F=−3100+120−3100\frac { 1 } { F } = \frac { - 3 } { 100 } + \frac { 1 } { 20 } - \frac { 3 } { 100 }

⇒ F=−100 cmF = - 100 \mathrm {~cm}