Download Solutions to Problems: Diffraction and Polarization

Solutions to Problems: Diffraction and Polarization P726: 7, 16, 19, 32, 36, 41, 43, 45, 51 7. How wide is the central diffraction peak on a screen 3.50 m behind a 0.0655‐mm‐wide slit illuminated by 400‐nm light? Solution: The angular position of first order dark fringe can be found as 1

sin  dark

a

0.4
 0.0061 , 65.5
so the width of the central diffraction peak is 1
w  2 f sin  dark
2f

a
 7.0 m 
0.4
 0.043 m . 65.5
16. How many fringes are contained in the central diffraction peak for a double‐slit pattern if (a) d = 2.00a, (b) d = 12.0 a, (c) d = 4.50 a, (d) d = 7.20a. Solution: The half angular width of the central diffraction reads 1
sin dark
  a. (a) For d = 2.00a, one can find d sin dark  2 , so there are 3 bright fringes 1
( sin bright  0,  d ) in the central diffraction peak. Similar to (a), one can figure out, (b) there are 2*(12‐1) + 1 = 23 bright fringes in the central diffraction, (c) there are 2*(5‐1) +1 = 9 bright fringes in the central diffraction, and (d) there are 2*(8‐1) +1 = 15 bright fringes in the central diffraction. 19. What is the angular resolution limit set by diffraction for the 100‐inch (mirror diameter) Mt. Wilson telescope (λ = 500 nm)? Solution: The limiting angle can be obtained as  min  1.22  D  1.22  5  105  2.54  100   2.40  107 . 32. The first‐order line of 589‐nm light falling on a diffraction grating is observed at a 15.5º angle. How far apart are the slits? At what angle will the third order be
observed?
Solution: According to the grating equation, we have
d sin15.5  589 nm  d  0.589 μm sin15.5  2.20 μm . The 3rd order line occurs at b3  sin 1  3 d   53.3 .
36. A 6500‐line/cm diffraction grating is 3.61 cm wide. If light with wavelengths near 624 nm falls on the grating, how close can two wavelengths be if they are to be resolved in any order? What order gives the best resolution? Solution: The grating constant and the number of the slits are d  1 cm 6500  1.538 μm, N  3.61  6500  23465. The maximum order m can be found as m  Int  d    2 , and the minimum separation between two wavelength near 624 nm is:    R   N  0.0266 nm , for m = 1, and    R   2 N  0.0133 nm , for m = 2. And the second order gives the best resolution. 41. First‐order Bragg diffraction is observed at 26.2º from a crystal with spacing between atoms of 0.24 nm. (a) At what angle will second order be observed? (b) What is the wavelength of the X‐rays? Solution: (a) According to Bragg’s law, one has 2d sin 26.2   and  2d sin  2  2 , which tells us  2  sin 1  2sin 26.2   62.0 . (b) The wavelength of the X‐ray reads   2d sin 26.2  0.212 nm . 43. Two polarizers are oriented at 75º to one another. Unpolarized light falls on them. What fraction of the light intensity is transmitted? Solution: The intensity of the transmitted light is I
1
I 0 cos2 75  0.0335I 0 , 2
thus, there are 3.35% of the light intensity is transmitted. 45. What is Brewster’s angle for an air‐glass (n = 1.56) surface? Solution: According to Brewster’s law, the angle reads   tan 1 1.56  57.3 . 51. Unpolarized light passes through five successive Polaroid sheets each of whose axis makes a 45º angle with the previous one. What is the intensity of the transmitted beam? Solution: The intensity of the transmitted beam reads I
4
1
1
I0 . I 0  cos2 45  
2
32