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  4. The power of a lens (biconvex) is 1.25 m -1 in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 cm and 40 cm respectively. The refractive index of surrounding medium :

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Q. The power of a lens (biconvex) is $1.25 \,m ^{-1}$ in particular medium. Refractive index of the lens is $1.5$ and radii of curvature are $20 \,cm$ and $40 \,cm$ respectively. The refractive index of surrounding medium :

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Solution:

$P =\frac{\mu_2}{ f }=\left(\mu_1-\mu_2\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)$
(For this formula refer to NCERT Part-2, Chapter-9, Page no. 328, solved example 8) $\left(\mu_1\right.$ is refractive index of lens and $\mu_2$ is of surrounding medium)
$ 1.25=\left(1.5-\mu_2\right)\left(\frac{1}{0.2}+\frac{1}{0.4}\right) $
$ \frac{1.25 \times 0.08}{0.6}=\left(1.5-\mu_2\right) $
$\Rightarrow \mu_2=\frac{4}{3}$