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 (μ2) = 3/2
Refractive index of the air (μ1) = 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
μ1 = 3/2
μ2 = 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.
