Question

A bichromatic source produces light having wavelengths in vacuum of 450 nm and 650 nm. The indices of refraction are 1.440 and 1.420, respectively. In the situation above, a ray of the bichromatic light, in air, is incident upon the oil at an angle of incidence of 50.0°. The angle of dispersion between the two refracted rays in the oil is closest to:

Answer 1

The angle of dispersion between the two refracted rays in the oil is 0.51°.

Step-by-step explanation:

Given that,

Wavelength in vacuum

`\lambda_(1)= 450\ nm`

`\lambda_(2)=650\ nm`

The indices of refraction is

`n_(1)=1.440`

`n_(2)=1.420`

We need to calculate the refracted angle for 450 wavelength

Using Snell's law

`(\sin\theta_(1))/(\sin\theta_(2))=(n_(2))/(n_(1))`

`\sin\theta_(2)=(n_(1)\sin\theta_(1))/(n_(2))`

`\theta_(2)=\sin^(-1)((n_(1)\sin\theta_(1))/(n_(2)))`

Put the value into the formula

`\theta_(2)=\sin^(-1)((1*\sin50)/(1.440))`

`\theta_(2)=32.13^(\circ)`

We need to calculate the refracted angle for 650 wavelength

Using Snell's law

`(\sin\theta_(1))/(\sin\theta_(2))=(n_(2))/(n_(1))`

`\theta_(2)=\sin^(-1)((n_(1)\sin\theta_(1))/(n_(2)))`

Put the value into the formula

`\theta_(2)'=\sin^(-1)((1*\sin50)/(1.420))`

`\theta_(2)'=32.64^(\circ)`

We need to calculate the angle of dispersion between the two refracted rays in the oil is

`\theta=\theta_(2)'-\theta_(2)`

Put the value into the formula

`\theta=32.64-32.13`

`\theta=0.51^(\circ)`

Hence, The angle of dispersion between the two refracted rays in the oil is 0.51°.