Question

Situation 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

Answer:

$\theta=0.52^{\circ}$

Explanation:

It is given that,

Wavelength in vacuum, $\lambda_1=450\ nm$

Wavelength in vacuum, $\lambda_2=650\ nm$

Refractive index for air, $n_1=1$

First refractive index, $n =1.44$

Second refractive index, $n' =1.42$

A ray is incident at an angle of incidence of 50 degrees. Let r is the angle of refraction. Firstly calculating the angle of refraction for two values of wavelength from Snell's law as :

$\dfrac{sin\ i}{sin\ r}=\dfrac{n_2}{n_1}$

$r=sin^{-1}(\dfrac{n_1\ sin\ i}{n})$

For 450 nm, $r=sin^{-1}(\dfrac{1\ sin(50)}{1.44})$

r = 32.13 degrees

For 650 nm, $r'=sin^{-1}(\dfrac{1\ sin(50)}{1.42})$

r' = 32.65 degrees

Let $\theta$ is the angle of dispersion between the two refracted rays in the oil such that,

$\theta=r'-r$

$\theta=32.65-32.13$

$\theta=0.52^{\circ}$

So, the angle of dispersion between the two refracted rays in the oil is closest to 0.52 degrees.