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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.
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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