Ch.37 8 9 SP1 35 29 38 59

Answers
8 . THIS IS EASY ! Take a look at the following graphic:

What does this picture tell you? It should say that the first minima (m= 0 in eqn.37.6 ) are directly across from each slit. Use formula 37.6 and set m = 0. Solve for L. By the way, what value do you use for Ydark ?  Look at the above graphic and make your decision based on simple geometric (i.e. common sense!) analysis and the information given on the distance d  between the slits!

9.  Use the "far field" approximation of equations 37.5 and 37.6. With that said and done, we continue. Use Equation 37.6 since we are talking about a minimum:

ym = the transverse distance on the screen from the axis to the location of the mth minimum. For this problem we assume that m = 0 for the first minimum. Clearly L = 150 m and ym = 20 m. Solve for d.
 SP1. At a particular location in a Young's interference pattern, the intensity on the screen is 6.4% of  the maximum.
(a) What minimum phase difference ( in radians) between sources produces this result?
(b) Find the path difference r2 - r1  for 587.5 nm light.


(a) Solve for the angle when I = 0.064I0

(b):
.
Solve for,
.

 
35.

 

The initial ray i is shown entering the left slab. Analyze the two rays that emerge from the right slab. Ray 1, which is transmitted from i without reflection, has undergone no phase change. Ray 2 has undergone two reflections before emerging from the right slab. There is a 1800 phase change at each reflection. Thus the net effect of these two reflections is no phase change at all. Thus, the phase difference between ray 1 and ray 2 is determined only by the extra distance that ray 2 travels before emerging on the right. Thus we have for constructive interference (bright-transmitted light) the usual relationship:
.

Note that the index of refraction of the glass does not effect the problem because the extra distance that ray 2 travels in the air gap, where n = 1. Find the minimum distance d. What value of m does that correspond to ?

 
 29.(a)



The initial ray i divides into two rays, ray1 that reflects off the first interface, and ray 2 that reflects off the second interface. Ray 1 undergoes a 1800 phase change since n for oil is greater than n for air. Ray 2 experiences no phase change since n for water is less that that for oil. Thus we have the "unusual" relations for constructive interference:


Note that this expression is in terms of the wavelength in oil, which is given by the wavelength in a vacuum divided by the index of refraction:

where n = 1.45, the index of refraction of oil. Be careful.

Solve for the wavelength in free space for the values of m = 0 and m = 1. Then state which colors that those wavelengths correspond to. Hint: the wavelength for m = 0 is not in the visible range. Try m = 1. Go to Ch. 40 for the frequencies various colors.
(b) A quick solution is to evaluate the difference (white - color). In other words subtract from white (all frequencies) the color you obtained from part (a): (white - color)=? The remaining frequencies give the dominant colors in the transmitted light. Refer to Ch. 40 again. Locate the range of frequencies about the frequency you obtained in part (a). You will notice that there is a range of approximately one color about the central frequency you computed above. The colors outside this range give those in the transmitted light. Hint: The color in part (a) is green. (white - green) = (red + violet) = purple.
38. The air gap behaves differently than in problem 35 since we are looking at the reflected  rays , not the transmitted rays. Let's represent the air gap of thickness d to analyze the reflections.:

Refer also to fig. 37.16. Ray 1 undergoes a phase change of 1800 upon reflection. Ray 2 undergoes no phase change. A dark band occurs when there is destructive interference:

This is the "unusual " condition because of the 1800 phase change for ray 1. Note that the wavelength is for light in a vacuum since the gap between the glass is air. Calculate m at the position of maximum thickness on the left end using the value of d at that location. Note that there is destructive interference at the position of zero thickness on the right end since at that point ray 1 is exactly 1800 out phase with ray 2. Therefore you must add the number 1 to the value of m you compute. The air gap varies between zero at the right end and the maximum d at the left end. Therefore, the number of values of thickness d corresponding to destructive interference and hence the number of dark bands is m + 1.
59. Look at fig P37.59. Let L be the length of each diagonal. Thus:


Compute L. Now, the path difference between the reflected ray and the direct wave is:

Compute this path difference.
Now, the reflected wave undergoes a half-wavelength (1800)phase shift upon reflection as stated. The expressions for constructive and destructive interference are switched from "normal.":
For constructive,

or,

And for destructive,

Or,

Find the longest wavelengths possible in each case by choosing the correct value of m that gives that maximum wavelength.