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Question
Two concave lenses L1 and L2 are kept in contact with each other. If the space between the two lenses is filled with a material of smaller refractive index, the magnitude of the focal length of the combination
Options
becomes undefined
remains unchanged
increases
decreases.
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Solution
increases
The focal length of the combination will increase. For finding out the combination of lens we have the formula:
\[\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}\]
Where, F is the focal length for the combination
d is the separation between two lenses
Here, d = 0
\[\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}\]
\[ F = \frac{f_1 f_2}{f_1 + f_2}\]
Hence, the focal length will increase.
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