For red, yellow, and violet rays, the focal lengths of a convex...

Using the lens maker's formula, the focal length $f$ is related to the refractive index $\mu$ by:

$\dfrac{1}{f} = (\mu - 1)$\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)

Let $K = \left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right).$The refractive indices for violet, red, and yellow light can be written as:

$\mu_v - 1 = \dfrac{1}{K f_v}$

$\mu_r - 1 = \dfrac{1}{K f_r}$

$\mu_y - 1 = \dfrac{1}{K f_y}$

The dispersive power $\omega$of the material is given by:

$\omega = \dfrac{\mu_v - \mu_r}{\mu_y - 1} = \dfrac{(\mu_v - 1) - (\mu_r - 1)}{\mu_y - 1}$

Substituting the expressions for the refractive indices:

$\omega = \dfrac{\dfrac{1}{K f_v} - \dfrac{1}{K f_r}}{\dfrac{1}{K f_y}} = \dfrac{\dfrac{1}{f_v} - \dfrac{1}{f_r}}{\dfrac{1}{f_y}} = \dfrac{f_r - f_v}{f_r f_v}f_y$

Given $f_r = 100 \text{ cm},f_y = 98 \text{ cm},$and $f_v = 96 \text{ cm}$:

$\omega = \dfrac{100 - 96}{100 \times 96} \times98$

$\omega = \dfrac{4}{9600} \times98 = \dfrac{98}{2400} \approx0.04083$

Rounding off to three decimal places, we get $\omega \approx0.041$.