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In Fig. 33-47a, a light ray in an underlying material is incident at an angleon a boundary with water, and some of the light refracts into the water. There are two choices of the underlying material. For each, the angle of refractionversus the incident angleis given in Fig. 33-47b.The horizontal axis scale is set byθ1s=90°.Without calculation, determine whether the index of refraction of

(a) material1 and

(b) material2 is greater or less than the index of water(n=1.33).What is the index of refraction of

(c) material 1 And

(d) material 2?

Short Answer

Expert verified
  1. This implies that a light ray travels from a denser medium to a rarer medium.

Hence.n2>n

2. This implies that a light ray travels from a denser medium to a rarer medium.

Hence.n2>n

3.The value of n1 from the graph of material 1 is n1=1.9.

4.The value of n2 from the graph of material 2 isn1=1.432 .

Step by step solution

01

Given data

The index of refraction for water is,n=1.33.

02

Determining the concept

Using the equation of Snell’s law and comparing the values of incident angle and refraction angle, whether the refractive index is greater or smaller can be determined. To calculate the actual value, take the values of angles from the graph.

The formula is as follows:

n1sinθ1=n2sinθ2

Where,

θ1= angle of incidence.

θ2= angle of refraction.

n1 = index of refraction of the incident medium.

n2 = index of refraction of the refractive medium.

03

(a) Determining without calculation, whether the index of refraction of a material  is greater or less than the index of water(n=1.33)

From the graph, it is observed that when light travels from material 1 to water, the angle of incidenceθ1is less than the angle of refractionθ2in water.

This implies that a light ray travels from a denser medium to a rarer medium.

Hencen1>n.

04

(b) Determining without calculation, whether the index of refraction of a material  is greater or less than the index of water(n=1.33)

From the graph, it is observed that when light travels from material 2 to water, the angle of incidence θ1 is less than the angle of refraction θ2 in water.

This implies that a light ray travels from a denser medium to a rarer medium.

Hence n2>n.

05

 Step 5: (c) Determining what is the index of refraction of material 1

From the graph of material 1,

θ2=90°

And

θ1=45°

Now use Snell’s law:

n1sinθ1=n2sinθ2

Heren1is the refractive index of material 1,n2is the refractive index of water.

Substitute the values in the above expression, and we get,

n1sin90∘=1.33sin90∘n1=1.9

Hence,the value of n1 from the graph of material 1 isn1=1.9 .

06

(d) Determining what is the index of refraction of material 2

From the graph of material 2,

θ2=90∘

And,

θ1=68.25°

Now use Snell’s law:

n1sinθ1=n2sinθ2

Heren1is the refractive index of material 2,n2is the refractive index of water.

Substitute the values in the above expression, and we get,

n1sin68.25∘=1.33sin90∘n1=1.432

Hence, the value of n1 from the graph of material 2 is n1=1.432.

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Question: In the ray diagram of the Figure, where the angles are not drawn to scale, the ray is incident at the critical angle on the interface between materials 2 and 3. Angle ϕ=60.0o, and two of the indexes of refraction are n1=1.70and n2=1.60.(a) Find index of refraction n3and(b) Find angle θ. (c) Ifθ is decreased, does light refract into material 3 ?

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