The focal lengths of a convex lens for red, yellow and violet rays are 100 cm, 98 cm and 96 cm respectively. Find the dispersive power of the material of the lens.

#### Solution

Focal lengths of the convex lens:-

For red rays, \[f_r = 100 cm\]

For yellow rays, \[f_y = 98 cm\]

For violet rays, \[f_v = 96 cm\]

Let

\[\mu_r\] = Refractive index for the red colour

\[\mu_y\] = Refractive index for the yellow colour

\[\mu_v\] = Refractive index for the violet colour

Focal length of a lens \[\left( f \right)\] is given by

\[\frac{1}{f} = \left( \mu - 1 \right)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]

Here, \[\mu\] is the refractive index and R_{1} and R_{2} are the radii of curvatures of the lens.

Thus, we have

\[\left( \mu - 1 \right) = \frac{1}{f} \times \frac{1}{\left( \frac{1}{R_1} - \frac{1}{R_2} \right)}\]

\[ \Rightarrow \left( \mu - 1 \right) = \frac{k}{f} ...........\left[ k = \frac{1}{\left( \frac{1}{R_1} - \frac{1}{R_2} \right)} \right]\]

For red rays,

\[\mu_r - 1 = \frac{k}{100}\]

For yellow rays,

\[\mu_y - 1 = \frac{k}{98}\]

For violet rays,

\[\mu_v - 1 = \frac{k}{96}\]

Dispersive power (ω) is given by

\[\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}\]

Or, \[\omega = \frac{( \mu_v - 1) - ( \mu_r - 1)}{( \mu_y - 1)}\]

Substituting the values, we get

\[\omega = \frac{\frac{k}{96} - \frac{k}{100}}{\frac{k}{98}} = \frac{98 \times 4}{9600}\]

\[ \Rightarrow \omega = 0 . 0408\]

Thus, the dispersive power of the material of the lens is 0.0408.