Question
Two identical thin plano convex lenses of refractive index $$n$$ are silvered, one on the plane side and the other on the convex side. The ratio of their focal lengths is
A.
$$\frac{n}{{n - 1}}$$
B.
$$\frac{{n - 1}}{n}$$
C.
$$\frac{{n + 1}}{n}$$
D.
$$n$$
Answer :
$$\frac{n}{{n - 1}}$$
Solution :
$$\eqalign{ & \frac{1}{f} = \frac{2}{{{f_e}}} + \frac{2}{{{f_m}}} \cr & {f_m} = \frac{R}{2} \cr & {f_m} = 0 \cr & \frac{1}{{{f_e}}} = \frac{{n - 1}}{R} \cr & {f_e} = \frac{{n - 1}}{R} \cr & {f_2} = \frac{R}{{2n}} \cr & {f_1} = \frac{R}{{2\left( {n - 1} \right)}} \cr & {\text{Ratio}}\,\,\frac{{{f_1}}}{{{f_2}}} = \frac{n}{{n - 1}} \cr} $$
$$\eqalign{ & \frac{1}{f} = \frac{2}{{{f_e}}} + \frac{2}{{{f_m}}} \cr & {f_m} = \frac{R}{2} \cr & {f_m} = 0 \cr & \frac{1}{{{f_e}}} = \frac{{n - 1}}{R} \cr & {f_e} = \frac{{n - 1}}{R} \cr & {f_2} = \frac{R}{{2n}} \cr & {f_1} = \frac{R}{{2\left( {n - 1} \right)}} \cr & {\text{Ratio}}\,\,\frac{{{f_1}}}{{{f_2}}} = \frac{n}{{n - 1}} \cr} $$
