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Two light pulses are emitted simultaneously and hit a screen directly in front of them. If one light pulse passes through 6.2 m of ice on its way to the screen, what is the time difference between the arrival of the two pulses at the screen? (The index of refraction of ice is 1.309.)
Suppose the distance to the screen is D . The time it takes for the first pulse to arrive is:
t1 =The time of flight for the second pulse is:
t2 = +where D - 6.2m is the distance travelled in air and v ice = c/n = c/1.309 is the speed of light in the ice. The time difference is therefore:
|t||=||t2 - t1|
|=||2.705 x 10- 8 s - 12.067 x 10- 8 s|
|=||6.38 x 10- 9 sec||(4)|
A light ray of wavelength = 589 nm is incident on glass with an angle of incidence of 30 o . The index of refraction of glass is 1.52. a)What is the angle of refraction? b) What are the speed and wavelength of the light inside the glass?
|=||sin 30 o = 0.329|
|=||= x 589 x 10- 6 m|
|=||387 x 10- 9 m = 387 nm|
|v2||=||= = 2 x 108 m/s||(6)|
First calculate the angle of refraction inside the glass:
sin = sinThen use as the angle of incidence onto the boundary between the glass and the air on the other side, so the angle of refraction in the air is:
A fish lives at the bottom of a lake 10m deep filled with water whose index of refraction 1.52. a) At what angle relative to the normal must the fish look up towards the surface of the water in order to see a fisherman who is sitting on a distant shore? b)What is the closest that another fish living at the bottom of the lake can approach in order that the first fish can see it by looking towards the surface?
|sin||=||sin 90 o =|
= sin = =Thus
d = = 20 m x 1.52 x = 30 mNote: that both the fisherman and the other fish have approximately the same apparent position to the first fish. Fish therefore have a pretty strange overhead view: if they look almost directly up, they see birds and clouds. If they look near the critical angle they see fisherman and trees on the distant shore. If they look at even steeper angles relative to the normal, they see other fish swimming ``overhead". I would like to sit in on an aquatic astronomer's conference to see how they explain it.
White light approaches one side of a glass prism at an angle of incidence of 40 o as shown in the diagram below. The apex angle of the prism is 60 o .
First calculate the angle of refraction inside the prism.
sin = sinFrom Table 22.1, we find that red light has an index of refraction: n2R = 1.51 , while for blue light, n2 = 1.53 . Thus:
|sin = sin 40|
|= 24.8 o|
|(90 o - ) + (90 o - ) = 180 o||(9)|
|= 60 o -||(10)|
= 60 o - 25.2 o = 34.8 oand
= 60 o - 24.8 o = 35.2 oFinally we calculate the angle of refraction sin of the emerging rays:
|sin = sin||(11)|
|=||(1.51 sin 34.8 o ) = 59.5 o|
|=||(1.53 sin 34.8 o ) = 61.8 o|
Monochromatic light goes through two thin slits 0.03mm apart. It is found that the second bright fringe on a screen 1.2 m away is 4.5 cm from the center. What is the wavelength of the light?
y2 = nwhere n = 2 , we find:
= = = 5.6 x 10- 7m = 560 nm
Note: This problem shows that is possible to measure very short wavelengths using the double slit experiment quite accurately as long as L/d is very large.
A radio transmitter transmits signals to a receiver 500m away along two paths. One is directly from the transmitter to the receiver, the other is bounced off a mountain a distance x directly behind the receiver. (See Fig. (22.16))
Since the rays bounce off a more dense mountain, there is a phase change on reflection, so the condition for destructive interference is:
r1 - r2 = 2x = (m + )The smallest value for x occurs when m = 0 , in which case:
x = = 75 m