Ray Optics

JEE Main level, single correct, medium

190 of 295

Question 190 of 295

A thin lens made of glass (refractive index =1.5=1.5 ) of focal length f=16cmf=16cm is immersed in a liquid of refractive index 1.421.42 . If its focal length in liquid is fl{f}_{l} , then the ratio fl/f{f}_{l}/f is closest to the integer:

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Answer: (B) 99

1fa=(μgμa−1)(1R1−1R2)\dfrac{1}{{f}_{a}}=\left(\dfrac{{\mu }_{g}}{{\mu }_{a}}-1\right)\left(\dfrac{1}{{R}_{1}}-\dfrac{1}{{R}_{2}}\right)
1fm=(μgμm−1)(1R1−1R2)\dfrac{1}{{f}_{m}}=\left(\dfrac{{\mu }_{g}}{{\mu }_{m}}-1\right)\left(\dfrac{1}{{R}_{1}}-\dfrac{1}{{R}_{2}}\right)
⇒fafm=(μgμm−1)(μgμa−1)=(1.501.42−1)[1.501−1]=0.08(1.92)(0.5)\Rightarrow \dfrac{{f}_{a}}{{f}_{m}}=\dfrac{\left(\dfrac{{\mu }_{g}}{{\mu }_{m}}-1\right)}{\left(\dfrac{{\mu }_{g}}{{\mu }_{a}}-1\right)}=\dfrac{\left(\dfrac{1.50}{1.42}-1\right)}{\left[\dfrac{1.50}{1}-1\right]}=\dfrac{0.08}{\left(1.92\right)\left(0.5\right)}
fafm=(1.42)(0.05)0.08=8.875≈9\dfrac{{f}_{a}}{{f}_{m}}=\dfrac{\left(1.42\right)\left(0.05\right)}{0.08}=8.875\approx 9

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