Two identical glass (μg=3/2) equiconvex lenses of focal length f are kept in contact. The space between the two lenses is filled with water (μw=4/3). The focal length of the combination is
A
f
B
f2
C
4f3
D
3f4
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Solution
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Let R be the radius of curvature of each surface. Then
1F=(1.5−1)(1R+1R)
For the water lens,
1f′=(43−1)(−1R+1R)=13(−2f);1f′=−23f
Now, using 1F=1f1+1f2+1f3, we have
1F=1f+1f+1f′=2f−23f=43f
⇒F=3f4.
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