Question

One end of a cylindrical glass rod (μ = 1.5) of radius 1.0 cm is rounded in the shape of a hemisphere. The rod is immersed in water (μ = 4/3) and an object is placed in the water along the axis of the rod at a distance of 8.0 cm from the rounded edge. Locate the image of the object.

Answer 1

Given,
Radius (R) of the cylindrical rod = 1.0 cm
Refractive index (μ_g) of the rod = 1.5 = $\frac{3}{2}$
Refractive index (μ_w) of water = 4/3



We know:

$\frac{{\mu }_g}{v}-\frac{{\mu }_w}{u}=\frac{{\mu }_g-{\mu }_w}{R}$
As per the question, u = −8 cm.
Now,
$$\begin{gathered} \frac{3}{2v}-\left(-\frac{4}{3\times 8}\right)=\frac{{\displaystyle \frac{3}{2}}-{\displaystyle \frac{4}{3}}}{1} \\ \Rightarrow \frac{3}{2v}+\frac{1}{6}=\frac{1}{6} \\ \Rightarrow v=\infty \end{gathered}$$

Hence, the image will be formed at infinity (∞).