JEE Main 12 Jan 2019 Evening Question 24
Question: A plano-convex lens (focal lengthy $ f _2 $ , refractive index $ {\mu_2}, $ radius of curvature R) fits exactly into a plano-concave lens (focal length $ f _1, $ refractive index $ {\mu_1}, $ radius of curvature -R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ f _1+f _2 $
B) $ \frac{R}{{\mu_2}-{\mu_1}} $
C) $ f _1-f _2 $
D) $ \frac{2f _1f _2}{f _1+f _2} $
Show Answer
Answer:
Correct Answer: B
Solution:
The focal length of the two lens is given
$ \frac{1}{f _2}=({\mu_2}-1)( \frac{1}{R _2}-\frac{1}{\infty } )=\frac{{\mu_2}-1}{R _2} $ $ \frac{1}{f _1}=({\mu_1}-1)( \frac{1}{\infty }-\frac{1}{R _1} )=-\frac{{\mu_1}-1}{R _1} $
So, the focal length of the lens combination is $ \frac{f _2f _1}{f _2+f _1}=\frac{( \frac{R _2}{{\mu_1}-1} )( \frac{-R _1}{{\mu_1}-1} )}{\frac{R _2}{{\mu_2}-1}-\frac{R _1}{{\mu_1}-1}}=\frac{R}{{\mu_2}-{\mu_1}} $
