Optics Question 818
Question: A piano convex lens fits exactly into a piano concave lens. Their plane surface are parallel to each other. If the lenses are made of different materials of refractive indices $ {\mu _{1}} $ & $ {\mu _{2}} $ and R is the radius of curvature of the curved surface of the lenses, then focal length of combination is
Options:
A) $ \frac{R}{{\mu _{1}}-{\mu _{2}}} $
B) $ \frac{2R}{{\mu _{1}}-{\mu _{2}}} $
C) $ \frac{R}{2( {\mu _{1}}-{\mu _{2}} )} $
D) $ \frac{R}{2-( {\mu _{1}}+{\mu _{2}} )} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \frac{1}{F}=\frac{1}{f _{1}}+\frac{1}{f _{2}} $
$ \frac{1}{F}=( {\mu _{1}}-1 )( \frac{1}{\infty }+\frac{1}{R} )+( {\mu _{2}}-1 )( \frac{1}{-R}-\frac{1}{\infty } ) $
$ =\frac{{\mu _{1}}-{\mu _{2}}}{R} $
$ F=\frac{R}{{\mu _{1}}-{\mu _{2}}} $
BETA
