Question

The adjacent figure shows a thin plano - convex lens of refractive index ${\mu }_1$ and a thin plano - concave R of their curved surfaces. The thin concavo - convex lens of refractive index ${\mu }_3$ has radius of curvature R of both its surfaces. The concavo - convex lens is so placed in between the plano - convex and plano concave lenses that the plane surfaces are parallel to each other. The focal length of the combination is

• A. $\frac{R}{({\mu }_1+{\mu }_2-{\mu }_3-1)}$
• B. $\frac{R}{({\mu }_1+{\mu }_2+{\mu }_3)}$
• C. $\frac{R}{({\mu }_1-{\mu }_2)}$
• D. $\frac{R}{({\mu }_1-{\mu }_2-{\mu }_3-3)}$

Answer 1

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

$\frac{1}{f_1}=({\mu }_1-1)(\frac{1}{\infty }+\frac{1}{R})=\frac{{\mu }_1-1}{R}\Rightarrow \frac{1}{f_1}=({\mu }_3-1)(-\frac{1}{R}+\frac{1}{R})=0$
$\frac{1}{f_3}=({\mu }_1-1)(-\frac{1}{R}-\frac{1}{\infty })=-\frac{({\mu }_2-1)}{R}$
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=\frac{({\mu }_1-1)}{R}-\frac{({\mu }_2-1)}{R}=\frac{1}{R}({\mu }_1-{\mu }_2)\Rightarrow f=\frac{R}{({\mu }_1-{\mu }_2)}$