Exercise 3: (8 points) Diffraction
A set up is performed by using a vertical rectangular slit of width a = 0.2 mm placed away from a screen by
𝐷 = 2.5 𝑚. Given that the speed of light in vacuum is: 𝑐 = 3 × 108 𝑚/𝑠.
Parts 1, 2 and 3 are independent.
1) Two electromagnetic waves of frequencies: 1 = 2 × 1020 Hz and ν2 = 5 × 1014 Hz are sent normally
on the vertical slit.
1.1) Calculate the wavelengths of each radiation.
a
1.2) Calculate the ratio λ for each radiation.
1.3) Which radiation produces the diffraction phenomenon using the slit? Justify.
2) A monochromatic light of wavelength = 800 nm falls normally on the vertical thin slit. The diffraction
figure is produced on the vertical screen.
2.1) Calculate the angular width and the linear width of the central fringe.
2.2) Deduce the width of each bright fringe surrounding the central fringe.
2.3) Determine, relative to the center O of the central fringe, the positions of the nearest points that have
zero intensity.
2.4) Specify how should D or a be varied in order to obtain a wider diffraction pattern.
2.5) What would be the angular width if:
2.5.1) We move the screen away and parallel to itself by 50 cm from the slit?
4
2.5.2) The device is put in water of index of refraction 𝑛 = 3 ?
3) In this part, a point source (S) emits a white light of wavelengths 𝜆𝑣𝑖𝑜𝑙𝑒𝑡 = 400 𝑛𝑚 ≤ 𝜆 ≤ 𝜆𝑟𝑒𝑑 = 800 𝑛𝑚
falls normally on the vertical thin slit.
3.1) The central bright fringe at O is white. Justify.
3.2) Show that the width of the white central fringe is the same as the width of the central bright fringe
obtained by 𝜆𝑣𝑖𝑜𝑙𝑒𝑡 = 400 𝑛𝑚. Deduce its value.
3.3) The curves of document-2 show the variation of the light intensity with respect to the x position
along the screen using the two light waves of wavelengths 𝜆𝑣𝑖𝑜𝑙𝑒𝑡 = 400 𝑛𝑚 𝑎𝑛𝑑 𝜆𝑟𝑒𝑑 = 800 𝑛𝑚.
Match, with justification, each curve to its corresponding wavelength.
I
I
I
(a) (b)
x x
0 0
Document-2
Exercise 3 8 pts
1) 𝑐 3×108 ½
𝜆1 = 𝜈 = 2×1020 = 1.5 × 10−12 𝑚
𝑐 3×108
𝜆2 = 𝜈 = 5×1014 = 0.6 × 10−7 𝑚
2
2) 𝑎 0.2×10−3 ½
= 1.5×10−12 = 0.133 × 109
𝜆1
𝑎 0.2×10−3
= 0.6×10−7 = 0.2 × 104
𝜆2
3) 𝜆2 = 0.6 × 10−7 𝑚 will undergo diffraction since a and 𝜆2 of same order of magnitude. ½
1) 𝜆 ½
𝑠𝑖𝑛𝜃 = 𝑎 ≈ 𝜃for the first angle of refraction (𝜃is very small) then
𝜆 800×10−9
𝜃=𝑎= = 4 × 10−3 𝑟𝑎𝑑
0.2×10−3
𝛼 = 2 |𝜃1 | = 2 × 10−3 × 4 = 8 × 10−3 𝑚 ½
𝐿 = 𝛼𝐷 = 8 × 10−3 × 2.5 = 20 × 10−3 𝑚 ½
2) 𝐿 ½
width of each 𝐵𝐹 = 2 = 10 × 10−3 = 10−2 𝑚
𝑥
3) 𝑡𝑎𝑛𝜃 = 𝐷 = 𝜃 (𝜃is very small) +figure 1
𝜆 𝑥 𝜆𝐷 800×10−9 ×2.5
𝜃 = 𝑎 = 𝐷 𝑡ℎ𝑒𝑛 𝑥 = ± =± = 10−3 𝑚
𝑎 0.2×10−3
4) Wider diffraction means larger 𝐿 1
𝜆𝐷
𝐿=2 𝐿 and 𝐷 are directly proportional then L increases by increasing D
𝑎
𝐿 and 𝑎 are inversely proportional then L increases by decreasing 𝑎
5) 𝜆 ½
5.1) 𝛼 = 2 it does not depend on D then it will stay the same
𝑎
𝜆 𝜆𝑎𝑖𝑟 8 ½
5.2) 𝛼 = 2 =𝛼=2 = 1.33 × 10−4 𝑚 = 6 × 10−4 𝑚
𝑎 𝑎𝑛
1) Each monochromatic light undergoes diffraction when it crosses the slit, and the center of the ½
C.B.F of each of these lights is at O. Constructive superposition of all lights takes place at O,
and then the color at O is white.
2) Since 𝜆𝑣𝑖𝑜𝑙𝑒𝑡 is the smallest , then 𝐿𝑣𝑖𝑜𝑙𝑒𝑡 is the smallest (𝐿 𝑎𝑛𝑑 𝜆 are proportional ½
The CBF appears white in the common area of the central fringes of all monochromatic light
radiations. Therefore, the width of the violet light represents the width of the central bright
fringe of light of the intersection of all light rays.
2(400×10−9 )2.5
𝐿= = 10−2 𝑚
0.2×10−3
3) 𝜆𝑟𝑒𝑑 > 𝜆𝑣𝑖𝑜𝑙𝑒𝑡 𝑡ℎ𝑒𝑛 𝐿𝑟𝑒𝑑 > 𝐿𝑣𝑖𝑜𝑙𝑒𝑡 figure (a) represents the pattern for 𝜆𝑣𝑖𝑜𝑙𝑒𝑡 ½
