Question

An achromatic doublet of focal length 90 cm is to be made of two lens. The material of one of the lense
has 1.5 times the dispersive power of the other. The doublet is converging type. Find the focal length
of each lens.​

Answer 1

The focal length  of each lens are 30 cm and -45 cm respectively.

Explanation:

The the effective focal length of the two lens, f = 90 cm.

For an achromatic combination,  

$\frac{\omega_1}{\omega_2} =-\frac{f_1}{f_2}$

Here $f_1,f_2$ are the focal lengths of the lenses and $\omega_1, \omega_2$ are the dispersive powers of the lenses.

Given, $\omega_2=1.5\omega_1$

Therefore,

$\frac{\omega_1}{1.5\omega_1}= -\frac{f_1}{f_2}$

$\frac{2}{3} =-\frac{f_1}{f_2}$.

Now the effective focal length of the lenses,

$\frac{1}{f} =\frac{1}{f_1} +\frac{1}{f_2}$                           (1)

substitute the values, we get

$\frac{1}{90cm} =\frac{1}{f_1}-\frac{2}{3f_1}$

$\frac{1}{90cm} =\frac{1}{3f_1}$

$f_1=30cm$

From equation (1), we get

$\frac{1}{90cm}=\frac{1}{30cm}+\frac{1}{f_2}$

$f_2=-45cm$

Thus, the focal length  of each lens are 30 cm and -45 cm respectively.

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