Question

The dispersive powers of the materials of the two lenses are in the ratio $3 : 4$. If the achromatic combination of these two lenses is of So focal length $60\, cm$, then the focal lengths of the component lenses are:

• A. +15 cm and - 20 cm
• B. - 15 cm and -20 cm
• C. +20 cm and -25 cm
• D. -20 cm and +25 cm

Answer 1

Answer: (A)

Let focal length of the convex and concavelenses be $f_{1}$ and $f_{2}$.
In order that this is achromatic combination of lenises, we have
$\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{F}$
Given, $F=60\, cm$
$\therefore \frac{1}{60}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\,\,\,\, ...(i)$
Also, condition for lens combination to be achromatic, we have
$\frac{\omega_{1}}{f_{1}}+\frac{\omega_{2}}{f_{2}} =0 $
$\Rightarrow \frac{\omega_{1}}{\omega_{2}} =-\frac{f_{1}}{f_{2}}$
where, $\omega_{1}$ and $\omega_{2}$ are dispersive powers.
we have given, $\frac{\omega_{1}}{\omega_{2}}=\frac{3}{4}$
$\therefore \frac{3}{4}=-\frac{f_{1}}{f_{2}}$
$\Rightarrow f_{2}=-\frac{4}{3} f_{1}\,\,\, ...(ii)$
Putting this value in (i), we have
$\frac{1}{60} =\frac{1}{f_{1}}-\frac{3}{4 f_{1}}$
$\Rightarrow f_{1} =+15 cm $
$f_{2} =-20 \,cm$
Hence, one is a convex lens and the other is a concave lens.