The refractive index of flint glass for blue | vedclass

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

Question

(A) The dispersive power $\omega$ is given by the formula $\omega = \frac{\mu_V - \mu_R}{\mu_y - 1}$.Given: $\mu_V = 1.6333$ (blue) and $\mu_R = 1.616

Vedclass · Accepted Answer

$0.0276$

Vedclass · Answer

(A) The dispersive power $\omega$ is given by the formula $\omega = \frac{\mu_V - \mu_R}{\mu_y - 1}$.Given: $\mu_V = 1.6333$ (blue) and $\mu_R = 1.6161$ (red).The mean refractive index $\mu_y$ is calculated as $\mu_y = \frac{\mu_V + \mu_R}{2} = \frac{1.6333 + 1.6161}{2} = 1.6247$.Now,substitute the values into the formula:$\omega = \frac{1.6333 - 1.6161}{1.6247 - 1} = \frac{0.0172}{0.6247} \approx 0.02753$.Rounding to the nearest provided option,we get $\omega = 0.0276$.

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.

Similar Questions

Calculate the dispersive power for crown glass from the given data if $\mu_v = 1.5230$ and $\mu_r = 1.5145$.

$A$ convex lens,a glass slab,a glass prism,and a solid sphere are all made of the same glass. The dispersive power will be:

The splitting of white light into several colours on passing through a glass prism is due to

What is dispersion of light?

Two incident monochromatic waves whose wavelengths differ by a small amount $d\lambda$ are separated angularly at $\theta$ and $\theta + d\theta$ with respect to the incident ray. The dispersive power is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module