Question

One end of a horizontal cylindrical glass rod $(\mu =1.5)$ of radius 5.0 cm is rounded in the shape of a hemisphere. An object 0.5 mm high is placed perpendicular to the axis of the rod at a distance of 20.0 cm from the rounded edge. Locate the image of the object and find its height.

• A. Same height as that of object & erect.
• B. Same height as that of object & inverted.
• C. Double the height of the object & inverted.
• D. Half the height of the object & inverted.

Answer 1

The correct option is B

Same height as that of object & inverted.

Taking the origin at the vertex,u = -20.0 cm and R = 5.0 cm.
We have,
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}or,\frac{1.5}{v}=\frac{1}{-20.0cm}+\frac{0.5}{5.0cm}=\frac{1}{20cm}or,v=30cm.$

The image is formed inside the rod at a distance of 30 cm from the vertex.
The magnification is $m=\frac{{\mu }_{1}v}{{\mu }_{2}u}$
$=\frac{30cm}{-1.5\times 20cm}=-1$.
Thus, the image will be of same height (0.5 mm) as the object but it will be inverted.