Physics 214 Spring 99|Problem Set 6|Solutions

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Week Beginning March 8: Serway 36.1, 36.2, 36.3, 36.4, 37.1, 37.4. 2. Serway, Chapter 34 ... The radiation pressure exerted by an electromagnetic wave when.
Physics 214 Spring 99|Problem Set 6|Solutions Handout March 11 1999

1. Reading Assignment

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(b). The total energy is a pulse will be given by the power times the duration of the pulse, U = P t = 25 kW  10,9 s = 25 J. (c). The average energy density in the pulse is the energy divided by the volume of the pulse, hui = U=V . The pulses, emerging in a parallel beam from the parabolic re ector, have an area A = R2 = (6 cm)2 = 113 cm2 . The length of each pulse is x = ct = 3  108 m=s  10,9 s = 30 cm. So the total volume of the pulse is V = Ax = 113  30 cm3 = 3390 cm3 . Then 25 J=V = 7:37  10,3 J=m3 hui = 3390 cm3 (d). p The peak electric eld is related to the average energy density by E = 2hui=0 . So, using 0 = 8:85  10,12 C2 =N , m2 , we have

Reading Assignment:

Week Beginning March 1: Serway 34.3, 34.4, 35.3, 35.4, 35.5, 35.6, 35.7 Week Beginning March 8: Serway 36.1, 36.2, 36.3, 36.4, 37.1, 37.4

2. Serway, Chapter 34, pg 1016, Problem 20 SOLUTION: The intensity of the light at 1 m will be given by I = P=A, where P =power, and A = 4r2 , the area of a sphere of radius r=1 m. Hence W 2 I = 4100 (1 m)2 = 7:96 W=m : The total average energy density is given by hui = I=c. Hence 2 hui = 37:9610W8 =mm=s = 2:65  10,8 J=m3 (a). The energy density is split equally into energy density in the electric eld and energy density in the magnetic eld. So huE i = hui=2 = 1:33  10,8 J=m3 ; and (b). huB i = hui=2 = 1:33  10,8 J=m3 : (c). The wave intensity, from above, is I = 7:96 W=m2 .

s

37  10,3 J=m3 = 4:08  104 V=m E = 8:2857:10 ,12 C2 =N , m2 The magnetic eld is

 104 V=m = 1:36  10,4 T: B = Ec = 4:08 3  108 m=s (e). The radiation pressure exerted by an electromagnetic wave when it is absorbed by a surface is p = I=c = hui. The total force will be

F = pA = huiA = 7:37  10,3 J=m3  113  10,4 m2 = 8:33  10,5 N:

4. Serway, Chapter 35, pg 1045, Problem 10

3. Serway, Chapter 34, pg 1016, Problem 46

SOLUTION: From Snell's Law, we have sin 2 = n1 =n2 sin 1 , where n1 = nair = 1, and n2 = nwater = 1:33. So sin 2 = (1=1:333)  sin 35:0 = 0:430 2 = 25:5

SOLUTION: (a). The wavelength of the microwaves is given by 3  108 m=s = 1:5 cm  = c=f = 20  109 s,1

1

the ber. For light incident at angles less than 2 , the light will exit the ber. We need to relate 2 to the angle of incidence of the light on the ber face. From the gure below

The wavelength of a wave in medium 2, 2 , is related to its wavelength in medium 1, 1 , by the relation

2 = 1 v2 v1

in which v1 = c=n1 is the wave speed in medium 1, and v2 = c=n2 is the wave speed in medium 2. So

2 = 1 vv12 = 1 nn12

The wavelength of sodium yellow light in air is 1 = 589 nm. So the wavelength in water is 2 = 589 nm 1:133 = 443 nm we see that sin  = n1 sin 1 and sin p1 = cos2 2 . At the critical angle, sin 2 = n2 =n1 , and therefore cos 2 = n21 , pn2 =n1 . Plugging into Snell's law we have sin  = n1 sin 1 = n1 cos 2 = n21 , n22 as desired.

5. Critical Angles

A light guide consists of a glass core (index of refraction n1) surrounded by a coating (index of refraction n2 < n1 ). Suppose a beam of light enters the ber from air at an angle  with the ber axis as shown in the gure below.

b) Assume the glass and coating indices of refraction are 1.58 and 1.53, respectively, and calculate the value of this angle. SOLUTION: 23.2

6. Serway, Chapter 35, pg 1047, Problem 34 SOLUTION: Let the index of refraction of the cube be n, and take the index of air to be 1. The gure below shows a light ray from the newspaper, entering the cube at point A, and making angle  with the normal. It is refracted at an angle .

a) Show that the greatest possible value of ,1 for2 which a ray can be 2 1=2 propagated down the ber is given by  = sin (n1 , n2 )

SOLUTION: The critical angle for the light incident on the glass/coating interface is sin 2 = n2 =n1 . For light incident at angles greater than 2 , the rays will be totally internally re ected and the light will propagate in

2

SOLUTION: The ray entering at A makes an angle with the normal to the circular surface. The refracted ray makes an angle with the normal. This ray travels to B, where it makes an angle  with the normal to the surface. The refracted ray at B makes an angle  with the normal.

Eye

α

A

Glass cube α

β

B β

L

n θ

φ α

R

B θ

A φ

α

Newspaper

From Snell's Law, the angles  and  are related by sin  = n sin  The maximum value that the angle  can have is 90 , so we have sin   1=n. The light ray travels through the cube and strikes the vertical side of the cube at point B. At this point, it makes at an angle with the normal, where = 90 , . The refracted ray emerging from the vertical side, and traveling to the eye, makes an angle with the normal, where, from Snell's Law, sin = n sin : The eye will see the newspaper if the light at point B is not totally internally re ected. This corresponds to the condition  90 , or sin  1=n. Since sin = cos , we have cos   1=n. Combining this with the condition on sin  given above, we get

From Snell's Law, we have sin = n sin and sin  = n sin . To relate  and , we note that, from the geometry of the gure, + = where sin = L=R. So, we have sin  = n sin ( , ) and   = arcsin sinn Combining these gives





sin  = n sin , arcsin sinn



. Using L=5 cm, R=10 cm, we get sin = 1=2, = 30 . Then, with n=2, we have   1  sin  = 2 sin 30 , arcsin 2  2 = 2 sin (30 , 14:48 ) = 0:535 So  = arcsin 0:535 = 32:4 .

sin2  + cos2  = 1  1=n2 + 1=n2 : p 2 So p 2=n  1, n  2. The maximum value which n can have is nmax = 2. No common glass actually has an index of refraction this low. A more realistic problem would replace the glass cube with a cube of clear plastic.

7. Serway, Chapter 35, pg 1048, Problem 46 3