Ray Optics

JEE Main level, single correct, medium

177 of 295

Question 177 of 295

The power of a lens (biconvex) is 1.25m−11.25{m}^{-1} in particular medium. Refractive index of the lens is 1.51.5 and radii of curvature are 20cm20cm and 40cm40cm respectively. The refractive index of surrounding medium:

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Answer: (D) 43\dfrac{4}{3}

Using Len's maker formula,
P=1f=(μ1−μ2)(1R1−1R2)P=\dfrac{1}{f}=\left({\mu }_{1}-{\mu }_{2}\right)\left(\dfrac{1}{{R}_{1}}-\dfrac{1}{{R}_{2}}\right)
Where, μ1{\mu }_{1} is refractive index of lens and μ2{\mu }_{2} is of surrounding medium.
Putting the values,
⇒1.25=(1.5−μ2)(10.2+10.4)\Rightarrow 1.25=\left(1.5-{\mu }_{2}\right)\left(\dfrac{1}{0.2}+\dfrac{1}{0.4}\right)
⇒1.25×0.080.6=(1.5−μ2)\Rightarrow \dfrac{1.25\times 0.08}{0.6}=\left(1.5-{\mu }_{2}\right)
The refractive index of surrounding medium is μ2=43{\mu }_{2}=\dfrac{4}{3}.

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