Respuesta :
Answer:
[tex]\theta=0.52^{\circ}[/tex]
Explanation:
It is given that,
Wavelength in vacuum, [tex]\lambda_1=450\ nm[/tex]
Wavelength in vacuum, [tex]\lambda_2=650\ nm[/tex]
Refractive index for air, [tex]n_1=1[/tex]
First refractive index, [tex]n =1.44[/tex]
Second refractive index, [tex]n' =1.42[/tex]
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 :
[tex]\dfrac{sin\ i}{sin\ r}=\dfrac{n_2}{n_1}[/tex]
[tex]r=sin^{-1}(\dfrac{n_1\ sin\ i}{n})[/tex]
For 450 nm, [tex]r=sin^{-1}(\dfrac{1\ sin(50)}{1.44})[/tex]
r = 32.13 degrees
For 650 nm, [tex]r'=sin^{-1}(\dfrac{1\ sin(50)}{1.42})[/tex]
r' = 32.65 degrees
Let [tex]\theta[/tex] is the angle of dispersion between the two refracted rays in the oil such that,
[tex]\theta=r'-r[/tex]
[tex]\theta=32.65-32.13[/tex]
[tex]\theta=0.52^{\circ}[/tex]
So, the angle of dispersion between the two refracted rays in the oil is closest to 0.52 degrees.
