course Phy232
Assignment 17 Ch. 22 Problems: 35, 38, 50, 52, 5535.
Index of refraction=n n=1.38 What is the largest angle of incidence theta
Sin theta= nb/na
Sin theta= 1/1.38= 0.72 or 72 degrees
38. Glass plate 2.50mm thick, index of refraction=1.40, placed between a point source of light with wavelength=540nm (in vacuum) and screen. Distance from source to screen is 1.80cm. How many wavelengths between source and screen?
Seperation is 1.4x10^7nm of air 2.5x10^6 n=1.4
450/1.4=321.4nm
1.8x10^7nm/450nm/wavelength= 4.0x10^4 wavelength in air
2.5xx10^6nm/321.4nm/wavelength= 7.78x10^3 wavelength in glass.
Right ideas, but note that in order to interfere the light makes a round trip.
50. Prove that a quarter-wave plate also converts circularly polarized light to linearly polarized light.
Quarter-wave plate consists of a carefully adjusted thickness of a material such that the light associated with the larger index of refraction is retarded by 90 degrees in phase with respect to that associated with the smaller index. Material is cut so the optic axis is parallel to the front and back plates of the plate. Any linearly polarized light which strikes the plate will be divided into two components with different indices of refraction. One of the useful applications of this device is to convert linearly polarized light to circularly polarized light.
That states it but doesn't prove it. Hint: the light that passes through the plate will in some directions be 1/2 wavelength, therefore 180 deg, out of phase with the light reflected from the surface.
52. Show that time “t” required for the light to travel 1 to 2 is
T= sqrt y2^2+x^2 + sqrt y2^2+(1-x)^2 / c
(B): Take the derivative of “t” with respect to x.
T= sqrt y1^2+x^2 + sqrt Y2^2+(1-x)^2/ c
= Y1^2 +x^2 + Y2^2 + (1-x)^2
= y1 + x +Y2 + (1-x)
It seems clear that you are using the principle of least action, minimizing the required time by optimizing the function, using a first-derivative test. This is the correct approach. However you haven't included enough information for me to assess all the details.
55.
(A): Show that the angle of deviation is given by
Sin A+ g/2 =nsinA/2
Na Sin theta 2=nb sin theta b
The triangle is divided by 2 with each side having A/2. You have to add g to the right side because it has it in the picture and the left doesn’t.
(B): Sin A+g/2 =n sinA/2 A=60 degrees n=1.52
Sin 60+g/2= 1.52sin 60/2
Sin30+g=0.76
g=0.76-0.5
g= 0.26
(C): Regractive index= 1.61 for red light (700nm), 1.66 for violet (400nm) A=60.0
Sin60+g/2= 1.61sin60/2 Sin60+g/2=1.66Sin60/2
0.5+8=0.805 0.5+g=0.83
g=0.805-0.5 g=0.83=0.5
g=0.305 g= 0.33
You don't have sufficient explanation of your figure. It appears you are finding a common point of constructive or destructive interference for two different wavelengths of light through some sort of aperture or grating, and your work is certainly plausible.