Question

An achromatic doublet of focal length $90\, cm$ is to be made of two lens. The material of one having $1.5$ times the dispersive power of other The doublet is converging type find focal length of each lens:-

• A. $f _{1}=45\, cm\, f _{2}=-30\, cm$
• B. $f _{1}=30\, cm\, f _{2}=-45\, cm$
• C. $f _{1}=-45\, cm\, f _{2}=60\, cm$
• D. $f _{1}=45\, cm\, f _{2}=-60\, cm$

Answer 1

Answer: (B)

For achromatic doublet
$\frac{\omega_{1}}{ f _{1}}+\frac{\omega_{2}}{ f _{2}}=0$
$\Rightarrow \frac{ f _{1}}{ f _{2}}=-\frac{\omega_{ l }}{\omega_{2}}$
$\frac{f_{2}}{f_{1}}=-1.5$ ...(1)
$\frac{1}{ f _{1}}+\frac{1}{ f _{2}}=\frac{1}{90}$ ...(2)
Solving eq.(1) and eq. (2)
$f _{1}=30\, cm ; f _{2}=-45\, cm$