For a material,the refractive indices for red | vedclass

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(A) The dispersive power $(\omega)$ of a material is defined as the ratio of the angular dispersion for extreme colors to the mean deviation.It is giv

Vedclass · Accepted Answer

$0.2$

Vedclass · Answer

(A) The dispersive power $(\omega)$ of a material is defined as the ratio of the angular dispersion for extreme colors to the mean deviation.It is given by the formula: $\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$Given:Refractive index for violet light,$\mu_v = 1.64$Refractive index for red light,$\mu_r = 1.52$Refractive index for yellow light,$\mu_y = 1.60$Substituting the values in the formula:$\omega = \frac{1.64 - 1.52}{1.60 - 1}$$\omega = \frac{0.12}{0.60}$$\omega = 0.2$Thus,the dispersive power of the material is $0.2$.

For a material,the refractive indices for red,violet,and yellow color light are respectively $1.52, 1.64,$ and $1.60.$ The dispersive power of the material is:

Similar Questions

If the refractive indices of crown glass for red,yellow,and violet colours are $1.5140, 1.5170$,and $1.5318$ respectively,and for flint glass these are $1.6434, 1.6499$,and $1.6852$ respectively,then the dispersive powers for crown and flint glass are respectively:

Explain the refractive index for light of different colours.

The refractive index of flint glass for blue $F$ line is $1.6333$ and red $C$ line is $1.6161.$ If the refractive index for yellow $D$ line is $1.622,$ the dispersive power of the glass is

What is a dispersive medium?

$A$ parallel beam of white light falls on a convex lens. Images of blue,yellow,and red light are formed on the other side of the lens at distances of $0.20 \ m$,$0.205 \ m$,and $0.214 \ m$ respectively. The dispersive power of the material of the lens will be:

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