Ray Optics

JEE Main level, single correct, medium

164 of 295

Question 164 of 295

A biconvex lens of refractive index 1.51.5 has a focal length of 20cm20cm in air. Its focal length when immersed in a liquid of refractive index 1.61.6 will be:

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Answer: (B) −160cm-160cm

Given, the refractive index of the material of the lens is μ1=1.5{\mu }_{1}=1.5, the refractive index of the medium is μm=1.6{\mu }_{m}=1.6 and the focal length of the lens in air is fa=20cm{f}_{a}=20cm.
In air:
1fa=(μ1−1)[2R]\dfrac{1}{{f}_{a}}=\left({\mu }_{1}-1\right)\left[\dfrac{2}{R}\right]
In liquid:
1fm=(μ1μm−1)[2R]\dfrac{1}{{f}_{m}}=\left(\dfrac{{\mu }_{1}}{{\mu }_{m}}-1\right)\left[\dfrac{2}{R}\right]
The focal length as measured in air and that as measured inside the liquid medium are related by,
fmfa=(μ1−1)μm(μ1−μm)...(1)\dfrac{{f}_{m}}{{f}_{a}}=\dfrac{\left({\mu }_{1}-1\right){\mu }_{m}}{\left({\mu }_{1}-{\mu }_{m}\right)}...\left(1\right)
Equation (1) implies that
fm20=(1.5−1)1.6(1.5−1.6)⇒fm=−160cm\dfrac{{f}_{m}}{20}=\dfrac{(1.5-1)1.6}{(1.5-1.6)} \Rightarrow {f}_{m}=-160cm

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