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# 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. - Physics

Sum

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.

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#### 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 R1 and R2 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.

Concept: Lenses
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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 20 Dispersion and Spectra
Q 3 | Page 442
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