Answer:
[tex]27.26^{\circ}[/tex]
Step-by-step explanation:
We are given that
[tex]n_g=1.57[/tex]
[tex]n_w=1.33[/tex]
Angle of incidence,i=37.5 degree
We have to find the angle of refraction into the water made by ray wit the normal to the surface.
Refractive index of water,[tex]n_a=1[/tex]
By Snell's law
[tex]\frac{n_g}{n_a}=\frac{sin i}{sin r}[/tex]
[tex]sin r=\frac{n_a sini}{n_g}=\frac{1\times sin37.5}{1.57}[/tex]
[tex]sin r=0.388[/tex]
[tex]r=sin^{-1}(0.388)=22.83^{\circ}[/tex]
Again for glass water surface
Using Snell's law
Angle of incidence,i'=22.83 degree
[tex]\frac{n_w}{n_g}=\frac{sin i}{sin r}[/tex]
[tex]\frac{1.33}{1.57}=\frac{sin22.83}{sin r'}[/tex]
[tex]sin r'=\frac{1.57sin(22.83)}{1.33}=0.458[/tex]
[tex]r'=sin^{-1}(0.458)=27.26^{\circ}[/tex]
The angle that the ray refracted into the water makes with the normal to the surface is; r = 27.24Ā°
We are given;
Refractive index of glass; Ī·_g = 1.57
Refractive index of water; Ī·_w = 1.33
Angle of incidence; i = 37.5Ā°
First of all, in air;
Refractive index of air is; Ī·_a = 1
We will use snell's law to find the angle of refraction. Thus;
Ī·_g/Ī·_a = sin i/sin r
ā“ 1.57/1 = (sin 37.5)/sin r
sin r = 0.3877
r = sinā»Ā¹0.3877
r = 22.81Ā°
The angle of 22.81Ā° will now be the incident angle for the water medium. Thus;
Ī·_w/Ī·_g = sin i/sin r
1.33/1.57 = sin 22.81/sin r
sin r = 0.3877/0.8471
sin r = 0.4577
r = sinā»Ā¹0.4577
r = 27.24Ā°
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