Physics - Geometrical Optics Question with Solution | TestHub

Dispersive power is ratio of angular dispersion and mean deviation.

It can be written in terms of focal length as,

$\omega =\frac{f_r-f_b}{f_y}$

Here $f_r=100 \mathrm{cm}$

$f_b=96.8 \mathrm{cm}$

$f_y=\sqrt{f_b\times f_r}$

$=\sqrt{96.8\times 100}$

$\approx 98.34 \mathrm{cm}$

$\therefore$ Dispersive power

$\omega =\frac{f_r-f_b}{f_y}$

$=\frac{100-96.8}{98.34}$

$\approx 0.0325$
Hence, the dispersive power of the material of the lens is $0.0325$.