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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 at .Physics Question Image

BITSATBITSAT 2018

Solution:

$
\begin{array}{l}
\mu_{1}(\text { air })=1 \\
\mu_{2} \text { (glass) }=1.5 \\
u =-15 cm \\
R =30 cm \\
\frac{\mu_{2}-\mu_{1}}{1.5-1}=\frac{\mu_{2}}{ v }-\frac{\mu_{1}}{4} \\
\frac{1.5}{30}-\frac{1}{15} \\
v =-30
\end{array}
$
Ans. $30 cm$ towards the left.