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Question
A point source of light is placed at a distance of 2 f from a converging lens of focal length f. The intensity on the other side of the lens is maximum at a distance
Options
f
between f and 2 f
2 f
more than 2 f.
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Solution
2 f
Since the object is placed at 2 f, the image of the object will be formed at distance of 2 ffrom a converging lens.
It can also be shown from the lens formula:
\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]
Here, u = − 2 f and f = f
On putting the respective values we get:
\[\frac{1}{v} - \frac{1}{- 2f} = \frac{1}{f}\]
\[ \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{2f}\]
\[ = \frac{1}{2f}\]
Therefore, image distance v = 2 f
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