Question: Q. 3. An object is placed in front of convex lens made of glass. How does the image distance vary if the refractive index of the medium is increased in such a way that still it remains less than the glass?
$\mathrm{R}$ [CBSE SQP 2016]
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Solution:
Ans. From the lens maker’s formula, it is clear that $n_{21}$ decreases then focal length increase.
$\frac{1}{f}=\left(n_{21}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \quad\left(\because n_{21}=\frac{n_{2}}{n_{1}}\right)$
Here refractive index of the glass with respect to surrounding material decreases. Hence, focal length increases which will also increase the image distance.
[CBSE Marking Scheme 2016]
