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"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers are parallel. The index of refraction of each layer is given in the drawing. Identical rays of light are sent into the layers, and light zigzags through each layer, reflecting from the top and bottom surfaces. The index of refraction for air is nair = 1.00. For each layer, the ray of light has an angle of incidence of 72.8˚. For the cases in which total internal refection is possible from either the top or bottom surface of a layer, determine the amount by which the angle of incidence exceeds the critical angle."

Respuesta :

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

pristina

The angle of incidence of the ray when it enters material B is 63.06°.The angle of incidence of the ray between A and B = 55.08°

Total Internal Reflection

According to Snell's law; at the interface of two media

[tex]n_1\,sin\,\theta_1= n_2\,sin\,\theta_2\\\\sin\,\theta_2=\frac{1.4\times \,sin\,72.8 ^\circ}{1.5} =0.8915\\\\\implies \theta _2 = sin^{-1}(0.8915)=63.06^\circ[/tex]

When the ray hits interface with material A;

[tex]1.5\, sin\,63.1^\circ = 1.3 \times sin \,\theta_3\\sin\,\theta_3= \frac{1.5\, sin\,63.06^\circ}{1.3}=1.028[/tex]

Here, the sine of the angle has a value of more than 1.

So, there is total internal reflection here.

When the ray hits the Air-C boundary;

[tex]1.4\times \,sin\,72.8 ^\circ = 1\, \times sin \,\theta\\\implies sin \,\theta= 1.4\times \,sin\,72.8 ^\circ =1.33[/tex]

Here also, the sine of the angle has a value of more than 1. So, there is total internal reflection here.

In case of material A, applying Snell's law;

[tex]1.3\times sin\,72.8^\circ = 1\times sin \,\theta\\\implies sin \,\theta=1.3\times sin\,72.8^\circ=1.24[/tex]

Here also, there will be total internal reflection.

When the ray hits the A-B interface.

[tex]1.3\times sin\,72.8^\circ = 1.5\,sin\,\theta\\sin\,\theta=\frac{1.3\times sin\,72.8^\circ}{1.5} =0.82\\\theta = sin^{-1}(0.82)=55.084[/tex]

Now, when the ray hits the C-Air boundary;

[tex]1.4\times sin\,63.06^\circ = 1 \times sin\,\theta\\sin\,\theta =1.4\times sin\,63.06^\circ =1.24[/tex]

So, as the ray of light passes through the layers of materials A, B and C the boundary with air on top and bottom will go through total internal reflection.

Learn more about total internal reflection here:

https://brainly.com/question/12995100

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