Question

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?

Answer 1

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:

$$\begin{gathered} \frac{{\mathrm{\mu }}_2}{v}-\frac{{\mathrm{\mu }}_1}{u}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{R} \\ \Rightarrow \frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{v}-\frac{1}{0}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{\infty } \\ \Rightarrow v=0 \end{gathered}$$

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:

$$\begin{gathered} \frac{{\mathrm{\mu }}_2}{v}-\frac{{\mathrm{\mu }}_1}{u}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{R} \\ \frac{1}{v}-\frac{3}{2\times (-3)}=\frac{1-{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{-3} \\ \Rightarrow \frac{1}{v}+\frac{3}{2\times 3}=\frac{{\displaystyle 1}}{6} \\ \Rightarrow \frac{1}{v}=\frac{{\displaystyle 1}}{6}-\frac{1}{2} \\ \Rightarrow v=-3\mathrm{cm} \end{gathered}$$

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