Neet Solved Paper 2013 Question 39
Question: A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices $ {\mu_1} $ and $ {\mu_2} $ and -R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is
Options:
A) $ \frac{R}{2({\mu_1}+{\mu_2})} $
B) $ \frac{R}{2({\mu_1}-{\mu_2})} $
C) $ \frac{R}{({\mu_1}-{\mu_2})} $
D) $ \frac{2R}{({\mu_2}-{\mu_1})} $
Show Answer
Answer:
Correct Answer: C
Solution:
Focal length of the combination $ \frac{1}{f}=\frac{1}{f _1}+\frac{1}{f _2} $
(i) We have $ f _1=\frac{R}{({\mu_1}-1)} $ and $ f _2=\frac{R}{({\mu_2}-1)} $ or $ \frac{1}{f _1}=\frac{R}{({\mu_1}-1)}or\frac{1}{f _2}=-\frac{R}{({\mu_2}-1)} $
Putting these values in Eq. (i), we get $ \frac{1}{f _1}=\frac{({\mu_1}-1)}{R}-\frac{({\mu_2}-1)}{R} $ $ =\frac{[{\mu_1}-1-{\mu_2}+1]}{R}=\frac{{\mu_1}-{\mu_2}}{R} $
