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Question: Focal length of plano-convex lens is F cm. This is made up of a material with m<sub>g</sub> = 2. If ...

Focal length of plano-convex lens is F cm. This is made up of a material with mg = 2. If it is place in the liquid of refractive index m = 43\frac { 4 } { 3 }, then its focal length will be-

A

F/2

B

2F

C

F

D

None of these

Answer

2F

Explanation

Solution

In air; 1 F=(2−1)(1R1−1R2)\frac { 1 } { \mathrm {~F} } = ( 2 - 1 ) \left( \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right) …(i)

In liquid; 1 F′=(24/3−1)(1R1−1R2)\frac { 1 } { \mathrm {~F} ^ { \prime } } = \left( \frac { 2 } { 4 / 3 } - 1 \right) \left( \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right)

\ 1 F′=(32−1)(1R1−1R2)\frac { 1 } { \mathrm {~F} ^ { \prime } } = \left( \frac { 3 } { 2 } - 1 \right) \left( \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right)…(ii)

(i)/(ii), 1 F1 F′=(2−1)(1R1−1R2)(32−1)(1R1−1R2)\frac { \frac { 1 } { \mathrm {~F} } } { \frac { 1 } { \mathrm {~F} ^ { \prime } } } = \frac { ( 2 - 1 ) \left( \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right) } { \left( \frac { 3 } { 2 } - 1 \right) \left( \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right) }

\ F' = 2F