A paperweight in the form of a hemisphere of radius 3.0cm is used to hold down a printed page. An observer looks at the page vertically through the paperweight. At what height above the page will the printed letters near the centre appear to the observer?
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In the first refraction at A.
$\mu_2=\frac{3}{2}, \ \mu_1=1, \ \text{u}=0, \ \text{R}=\infty$
So, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\Rightarrow \frac{\frac{3}{2}}{\text{v}}-\frac{1}{0}=\frac{\mu_2-\mu_1}{\infty}$
$\Rightarrow\text{v}=0$ since $\big(\text{R}\Rightarrow\infty \ \text{and} \ \text{u}=0\big)$
$\therefore$ The image will be formed at the point, Now for the second refraction at B,
$\frac{\mu_2}{\text{v}}-\frac{\mu}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\text{u}=-3\text{cm}, \ \text{R}=-3\text{cm}, \ \mu_1=\frac{3}{2}, \ \mu_2=1$
So, $\frac{1}{\text{v}}+\frac{3}{2\times3}=\frac{1-1.5}{-3}=\frac{1}{6}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{2}=-\frac{1}{3}$
$\Rightarrow \text{v}=-3\text{cm},$
$\therefore$ There will be no shift in the final image.
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