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Fundamentals Of Physics
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Short Answer

In Fig. 35-34, a light ray is an incident at angle θ1=50°on a series of five transparent layers with parallel boundaries. For layers 1 and 3 , L1=20 μm , L2=25 μm , n1=1.6 and n3=1.45 . (a) At what angle does the light emerge back into air at the right? (b) How much time does the light take to travel through layer 3?

(a) The angle of immerge of the light into air is 50o.

(b) The time taken by the light to travel through medium 3 is 0.142 ps.

See the step by step solution

Step by Step Solution

Step 1: Write the given data from the question.

The incident angle, θ1=50°

The length of the medium 1, L1=20 μm

The length of the medium 3, role="math" localid="1663069065664" L3=25 μm

The refractive index of medium 1, n1=1.6

The refractive index of medium 3, n3=1.45

Step 2: Determine the required formulas:

Snell's law of refraction states that: The incident ray, the refracted ray, and the normal at the point of incidence all lie in the same plane. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a pair of given environments.

The expression for Snell’s law is given as follows.

nsinθ1=n1sinθ2=n2sinθ3.........

The expression to calculate the time is given as follows.

t=n3Lc .......(1)

Step 3: (a) Calculate the angle at which light emerges back into the air:

Let assume angle θ2,θ3,θ4,θ5 and θ6 are the refracted angle in first, second, third, fourth, fifth, and air respectively.

Use Snell’s law for the air and first medium interface.

nairsinθ1=n1sinθ2

Use the Snell’s law for the first and second medium.

n1sinθ2=n2sinθ3

Substitute nairsinθ1 for n2sinθ3 into above equation.

nairsinθ1=n3sinθ4

Use the Snell’s law for the second and third medium.

n2sinθ3=n3sinθ4

Use the Snell’s law for the third and fourth medium

n3sinθ4=n4sinθ5

Substitute nairsinθ1 for n3sinθ4 into above equation.

nairsinθ1=n4sinθ5

Use the Snell’s law for the fourth and fifth medium.

n4sinθ5=n5sinθ6

Substitute nairsinθ1 for n4sinθ5 into above equation.

nairsinθ1=n5sinθ6

Use the Snell’s law for the fifth and air medium.

n5sinθ6=nairsinθ7

Substitute nairsinθ1 for n5sinθ6 into above equation

nairsinθ1=nairsinθ7sinθ1=sinθ7θ1=θ7

Substitute 50° for θ1 into above equation.

θ7=50°

Therefore, the angle of immerge of the light into air is 50 degree.

Step 4: (b) Calculate the time taken by the light to travel through layer  :

Consider the distance travelled by the light in third medium.

Recall the equation (2),

nairsinθ1=n3sinθ4sinθ4=nairn3sinθ1

Substitute 1 for nair , 1.45 for θ3 and 50° for θ1 into above equation.

θ4=11.45sin50°sinθ4=0.528θ4=sin-10.528θ4=31.89°

From the figure,

θ4=L3LL=L3cosθ4

Substitute25 μm for L3 and 31.89° for θ4 into above equation.

L=25×10-6cos31.89°L=2.944×10-5 m

Calculate the time taken by the light to travel through medium 3.

Substitute 2.944×10-5 m for L , 1.45 for n3 and 3×108ms for c into equation (1).

t=1.45×2.944×10-53×108=4.2688×10-133=1.42×10-13=0.142 ps

Hence the time taken by the light to travel through medium 3 is 0.142 ps .

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