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 index ${\mu }_1$ and ${\mu }_2$ and R is the radius of curvature of the curved surfaces of the lenses, then the focal length of the combination is:

• A. $\frac{R}{(2{\mu }_1+{\mu }_2)}$
• B. $\frac{R}{2({\mu }_1-{\mu }_2)}$
• C. $\frac{R}{{\mu }_1-{\mu }_2}$
• D. $\frac{2R}{{\mu }_1-{\mu }_2}$

Answer 1

The correct option is C $\frac{R}{{\mu }_1-{\mu }_2}$

It is combination of two lenses 1 and 2 It is combination of two lenses 1 and 2
$\frac{1}{f_c}=\frac{1}{f_1}+\frac{1}{f_2}$
Using lens maker formula
$\frac{1}{f_1}=({\mu }_1-1)(\frac{1}{\infty }-\frac{1}{R}=+\left[\frac{{\mu }_1-1}{R}\right])\frac{1}{f_2}=({\mu }_2-1)(-\frac{1}{R}-\frac{1}{\infty })=-\left[\frac{{\mu }_2-1}{R}\right]\frac{1}{f_c}=\frac{{\mu }_1-1}{R}-\frac{({\mu }_2-1)}{R}\frac{1}{f_c}=\frac{{\mu }_1-1-{\mu }_2+1}{R}=\frac{{\mu }_1-{\mu }_2}{R}{f}_c=\frac{R}{{\mu }_1-{\mu }_2}$