A paperweight in the form of a hemisphere of radius 3.0 cm 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?

#### Solution

Given:

Radius of the paperweight (*R*) = 3 cm

Refractive index of the paperweight (μ_{2}) = 3/2

Refractive index of the air (μ_{1}) = 1

In the first case, the refraction is at A.*u =* 0 and *R =* ∞

We know:

\[\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\]

\[ \Rightarrow \frac{\frac{3}{2}}{v} - \frac{1}{0} = \frac{\mu_2 - \mu_1}{\infty}\]

\[ \Rightarrow v = 0\]

Therefore, the image of the letter is formed at the point.

For the second case, refraction is at point B.

Here,

Object distance, *u *= −3 cm*R *= − 3 cm

μ_{1} = 3/2

μ_{2} = 1

Thus, we have:

\[\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\]

\[\frac{1}{v} - \frac{3}{2 \times ( - 3)} = \frac{1 - \frac{3}{2}}{- 3}\]

\[ \Rightarrow \frac{1}{v} + \frac{3}{2 \times 3} = \frac{1}{6}\]

\[ \Rightarrow \frac{1}{v} = \frac{1}{6} - \frac{1}{2}\]

\[ \Rightarrow v = - 3 \text{ cm }\]

Hence, there will be no shift in the final image.