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QJEEM20Ph328e: JEE Main 2020 Physics - Class 12 JEEM20Ph328e Solution: A thin lens made of glass (refractive index = 1.5) of focal…

A thin lens made of glass (refractive index = 1.5) of focal length f = 16 cm is immersed in a liquid of refractive index 1.42. If its focal length in liquid is f1 , then the ratio f1f{{{f_1}} \over f} is closest to the integer :

Multiple Choice Options

  • A 17
  • B 1
  • C 9
  • D 5

Step-by-Step Verified Solution

Using formula
1f=(μ2μ1−1)(1R1−1R2){1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

1f=(1.51−1)(1R1−1R2){1 \over {{f}}} = \left( {{{1.5} \over 1} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) ...(1)

and 1f1=(1.51.42−1)(1R1−1R2){1 \over {{f_1}}} = \left( {{{1.5} \over {1.42}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) ...(2)

Dividing (1) by (2), we get

f1f=0.50.056{{{f_1}} \over f} = {{0.5} \over {0.056}} = 8.93 ≈\approx 9

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