How will you prove refractive inside of red is less than that of violet by formulas?

Question

Charles Aeschliman · Accepted Answer

This can be proved using Cauchy's equation which is stated above #n (lambda)#: refractive index of respective colour #lambda#: wavelength of the light ray From the equation we can see that refractive index decreases with increasing wavelength Hence as wavelength of red is more than violet Therefore refractive index of red is less than that of violet.

Charlotte Aaron · Answer

undefined You can prove that the refractive index of red light is less than that of violet light using Snell's Law, which relates the angle of incidence, angle of refraction, and refractive indices of two mediums. Snell's Law is given by:

$$ n_1 \times \sin(\theta_1) = n_2 \times \sin(\theta_2) $$

Where:
- $ n_1 $ and $ n_2 $ are the refractive indices of the two mediums.
- $ \theta_1 $ and $ \theta_2 $ are the angles of incidence and refraction, respectively.

Since the speed of light is higher in mediums with lower refractive indices, you can infer that if red light (with a longer wavelength) bends less than violet light (with a shorter wavelength) when passing through the same medium, the refractive index of red light is less than that of violet light.

This relationship is evident in Snell's Law when comparing the angles of incidence and refraction for red and violet light passing through the same medium. If $ \theta_1 $ and $ \theta_2 $ are smaller for red light compared to violet light, it implies that $ n_1 < n_2 $ for red light compared to violet light in that medium.