Respuesta :
Answer:
Explanation:
Lower the refractive index , higher the wave length
Refractive index becoming high reduces the velocity and hence wavelength as frequency remains unchanged.
So refractive index 1.4831 will correspond to wavelength of 450 nm.
For refractive index is 1.4831 , angle of incidence i , angle of refraction r .
Sin i / Sinr = 1.4831
sin65 / sinr = 1.4831
sir r = sin65 / 1.4831
= .9063 / 1.4831
= .6111
r = 37.67 degree
For refractive index is 1.4754 , angle of incidence i
Sin i / Sinr = 1.4754
sin65 / sinr = 1.4754
sir r = sin65 / 1.4754
= .9063 / 1.4754
= .61427
r = 37.9 degree
angle between two refracted ray
37.9 -37.67
= 0. 23 degree
Answer:
0.2°
Explanation:
Angle of incidence on glass, i = 65°
first wavelength, λ1 = 450 nm
second wavelength, λ1 = 700 nm
first refractive index, μ1 = 1.4831
second refractive index, μ1 = 1.4754
Let the angle of refraction for the first wavelength is r1 and for the second wavelength is r2.
Use Snell's law
[tex]\mu_{1}=\frac{Sin i}{Sin r_{1}}[/tex]
[tex]Sin r_{1}=\frac{Sin 65}{1.4831}[/tex]
Sin r1 = 0.61109
r1 = 37.7°
[tex]\mu_{2}=\frac{Sin i}{Sin r_{2}}[/tex]
[tex]Sin r_{2}=\frac{Sin 65}{1.4754}[/tex]
Sin r2 = 0.6143
r2 = 37.9°
Angle between the two refracted beams = r2 - r1 = 37.9° - 37.7° = 0.2°
