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Q. A point object $O$ is placed in front of a glass rod having spherical end of radius of curvature $30\, cm$. The image would be formed atPhysics Question Image

ManipalManipal 2019

Solution:

Focal length of glass rod, $f=\frac{R}{2}=\frac{30}{2}=15\, cm$
Distance of object, $u=-15\, cm$
From formula
$\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$
$\mu_{2}=1.5$ (for glass)
$\mu_{1}=1$ (for air)
$\frac{1.3}{v}-\frac{1}{-15}=\frac{1.5-1}{30}$
$v=-30\, cm$