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JEE Main 2016 · Physics · Ray and Wave Optics · hard

A thin biconvex glass lens of refractive index 1.5 has a focal length of 20 cm in air. When it is completely immersed in water of refractive index…

Single correct option

A thin biconvex glass lens of refractive index 1.51.5 has a focal length of 20 cm20\ \mathrm{cm} in air. When it is completely immersed in water of refractive index 43\dfrac43, its focal length becomes
  1. A. 40 cm40\ \mathrm{cm}
  2. B. 60 cm60\ \mathrm{cm}
  3. C. 80 cm80\ \mathrm{cm}
  4. D. 26.7 cm26.7\ \mathrm{cm}
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Answer: (C) 80 cm80\ \mathrm{cm}

The lens-maker formula in a surrounding medium is 1f=(μgμm−1)(1R1−1R2)\dfrac1f=\left(\dfrac{\mu_g}{\mu_m}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right). In air, 120=(1.5−1)K\dfrac1{20}=(1.5-1)K, so K=0.1 cm−1K=0.1\ \mathrm{cm^{-1}}. In water, μgμm=1.54/3=98\dfrac{\mu_g}{\mu_m}=\dfrac{1.5}{4/3}=\dfrac98, so μgμm−1=18\dfrac{\mu_g}{\mu_m}-1=\dfrac18. Then 1f′=18×0.1=180\dfrac1{f'}=\dfrac18\times0.1=\dfrac{1}{80}, giving f′=80 cmf'=80\ \mathrm{cm}. The focal length is four times as large because 0.50.125=4\dfrac{0.5}{0.125}=4. Using (μg−μm)(\mu_g-\mu_m) instead of the ratio would give 60 cm60\ \mathrm{cm}.
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