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A narrow parallel beam of light falls on a glass sphere of radius $\mathrm{R}$ and refractive index $\mu$ at normal incidence. The distance of the image from the outer edge is given by-
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Verified Answer
The correct answer is:
$\frac{\mathrm{R}(2-\mu)}{2(\mu-1)}$
now $\frac{\mu}{\mathrm{V}_{1}}-\frac{1}{\infty}=\frac{\mu-1}{\mathrm{R}} \Rightarrow \mathrm{V}_{1}=\frac{\mu \mathrm{R}}{\mu-1}$
now $\frac{1}{\mathrm{~V}_{\mathrm{f}}}-\frac{\mu}{-\left(2 \mathrm{R}-\mathrm{V}_{1}\right)}=\frac{1-\mu}{-\mathrm{R}}$
replace $\mathrm{V}_{1}$ by $\frac{\mu \mathrm{R}}{\mu-1}$ and ving for $\mathrm{V}_{\mathrm{f}}$
we get $\mathrm{V}_{\mathrm{f}}=\frac{\mathrm{R}(\mu-2)}{2(\mu-1)}$
First image is real and second is virtual.
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