Respuesta :
Answer:
The angle of the incident ray at the air-glass interface is 62.86°
Explanation:
Given that,
Refractive index of glass =1.60
Angle = 42°
We need to calculate the angle of the incident ray at glass-water interface
Using Snell's law
[tex]n_{2}\sin\theta_{2}=n_{3}\sin\theta_{3}[/tex]
Where, [tex]n_{2}[/tex] = refractive index of glass
[tex]n_{3}[/tex] = refractive index of water
[tex]\theta_{3}[/tex] = angle of refraction
Put the value into the formula
[tex]1.6\sin\theta_{2}=1.33\sin42[/tex]
[tex]\sin\theta_{2}=\dfrac{1.33\sin42}{1.6}[/tex]
[tex]\theta_{2}=\sin^{-1}(\dfrac{1.33\sin42}{1.6})[/tex]
[tex]\theta_{2}=33.8^{\circ}[/tex]
We need to calculate the angle of the incident ray at the air-glass interface
Using Snell's law
[tex]n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}[/tex]
Where, [tex]n_{1}[/tex] = refractive index of air
[tex]n_{2}[/tex] = refractive index of glass
[tex]\theta_{1}[/tex] = angle of incident
Put the value into the formula
[tex]1\sin\theta_{1}=1.6\sin33.8[/tex]
[tex]\theta=\sin^{-1}(1.6\sin33.8)[/tex]
[tex]\theta=62.86^{\circ}[/tex]
Hence, The angle of the incident ray at the air-glass interface is 62.86°
Answer:
62.8 degree
Explanation:
Let the incident ray incident at an angle [tex]\theta_1[/tex] at air glass surface.
[tex]\theta_3=42^{\circ}[/tex]=Angle of refraction when ray travel from glass to water
[tex]\theta_2=[/tex]Angle of refraction when the ray travel from air to glass
Refractive index of glass,[tex]n_2=1.6[/tex]
We know that
Refractive index of water=[tex]n_3=1.33[/tex]
Snell's law
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]
Where [tex]\theta_1[/tex]=Angle of incidence
[tex]\theta_2=[/tex]Angle of refraction
[tex]n_1=[/tex]Refractive index of medium 1
[tex]n_2=[/tex]Refractive index of medium 2
When the ray travel from glass to water
[tex]n_2sin\theta_2=n_3sin\theta_3[/tex]
Where [tex]n_2=[/tex]Refractive index of medium 1(Glass)
[tex]n_3[/tex]=Refractive index of medium 2 (Water)
[tex]\theta_2=[/tex]Angle of incidence
[tex]\theta_3[/tex]=Angle of refraction
Substitute the values
[tex]1.6sin\theta_2=1.33sin42[/tex]
[tex]sin\theta_2=\frac{1.33sin42}{1.6}[/tex]
[tex]sin\theta_2=0.556[/tex]
[tex]\theta_2=sin^{-1}(0.556)=33.8^{\circ}[/tex]
Angle of refraction when ray travel from air to glass= Angle of incidence of when ray travel from glass to water
Angle of refraction when the ray travel from air to glass=33.8 degree
Refractive index of air=[tex]n_1=1[/tex]
Again apply Snell's law
[tex]n_1sin\thet_1=n_2sin\theta_2[/tex]
[tex]1\times sin\theta_1=1.6sin(33.8)[/tex]
[tex]sin\theta_1=1.6\times 0.556=0.8896[/tex]
[tex]\theta_1=sin^{-1}(0.8896)=62.8^{\circ}[/tex]
Hence, the angle of the incident ray at the air-glass interface=62.8 degree
