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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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Most popular questions from this chapter

Suppose we rotate the second sheet in Fig. 33-15a, starting with the polarization direction aligned with the y axis(θ=00)and ending with it aligned with the x-axis(θ=90°). Which of the four curves in Fig. 33-26 best shows the intensity of the light through the three-sheet system during this90°rotation?

One want to rotate the direction of polarization of a beam of polarized light throughby sending the beam through one or more polarizing sheets.

(a) What is the minimum number of sheets required?

(b) What is the minimum number of sheets required if the transmitted intensity is to be more than60%of the original intensity?

In Figure

(a), a beam of light in a material1is incident on a boundary at an angle θ1=40°. Some of the light travels through the material 2, and then some of it emerges into the material 3. The two boundaries between the three materials are parallel. The final direction of the beam depends, in part, on the index of refraction n3of the third material. Figure (b) gives the angle of refraction θ3in that material versus n3a range of possiblen3values. The vertical axis scale is set byθ3a=30.0° and θ3b=50.0°.(a) What is the indexof refraction of material , or is the index impossible to calculate without more information?

(b) What is the index of refraction of material 2, or is the index impossible to calculate without more information?

(c) It θ1is changed to 70°and the index of refraction of a material 3 is2.4 , what is θ3?

Figure:

Figure 33-32 shows four long horizontal layers ´¡â€“D of different materials, with air above and below them. The index of refraction of each material is given. Rays of light are sent into the left end of each layer as shown. In which layer is there the possibility of totally trapping the light in that layer so that, after many reflections, all the light reaches the right end of the layer?

Each part of Fig. 33-34 shows light that refracts through an interface between two materials. The incident ray (shown gray in the figure) consists of red and blue light. The approximate index of refraction for visible light is indicated for each material. Which of the three parts show physically possible refraction? (Hint: First consider the refraction in general, regardless of the color, and then consider how red and blue light refract differently?

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