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QJEEM21Ph65d6: JEE Main 2021 Physics - Class 12 JEEM21Ph65d6 Solution: The refractive index of a converging lens is 1.4. What will be…

The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index? Assume the radii of curvature of the faces of lens are R1 and R2 respectively.

Multiple Choice Options

  • A Zero
  • B Infinite
  • C 1
  • D R1R2R1−R2{{{R_1}{R_2}} \over {{R_1} - {R_2}}}

Step-by-Step Verified Solution

Given, initially refractive index (n1) = 1.4

Then it placed in medium of same refractive index.

∴\therefore n2 = 1.4

We know, Focal length

1f=(n1n2−1)(1R1−1R2){1 \over f} = \left( {{{{n_1}} \over {{n_2}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

⇒\Rightarrow 1f=(1.41.4−1)(1R1−1R2){1 \over f} = \left( {{{1.4} \over {1.4}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

⇒\Rightarrow 1f=0{1 \over f} = 0

⇒\Rightarrow f = infinite

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