The refractive index of the material of the p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The dispersive power $(\omega)$ of a prism is defined as the ratio of the angular dispersion to the mean deviation.It is given by the formula: $\o

Vedclass · Accepted Answer

$0.06$

Vedclass · Answer

(B) The dispersive power $(\omega)$ of a prism is defined as the ratio of the angular dispersion to the mean deviation.It is given by the formula: $\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$Where:$\mu_v = 1.69$ (refractive index for violet colour)$\mu_r = 1.65$ (refractive index for red colour)$\mu_y = 1.66$ (refractive index for mean colour)Substituting the values into the formula:$\omega = \frac{1.69 - 1.65}{1.66 - 1}$$\omega = \frac{0.04}{0.66}$$\omega = \frac{4}{66} = \frac{2}{33} \approx 0.0606$Rounding to two decimal places,we get $\omega = 0.06$.

The refractive index of the material of the prism for violet colour is $1.69$ and that for red is $1.65$. If the refractive index for the mean colour is $1.66$,calculate the dispersive power of the material of the prism.

Similar Questions

The refractive index of flint glass for blue light is $1.6333$ and for red light is $1.6161$. Calculate the dispersive power of the prism,assuming the mean refractive index is approximately the average of the two given values.

In the phenomenon of dispersion without deviation,which of the following statements is true?

Mention the constituent colors of white light.

The dispersive power is maximum for the material

Compare the dispersive powers of two prisms if one of them deviates the blue and red rays through $10^{\circ}$ and $6^{\circ}$ respectively,and the second prism through $8^{\circ}$ and $4.5^{\circ}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module