Figure (17-E5) shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let
BP0-AP0=λ/3 and D>>λ. (a) Show that in this case
d=2λD/3. (b) Show that the intensity at P0 is three times the intensity due to any of the three slits individually.
Answers
Given:
Wavelength of light = λ
Path difference of wave fronts reaching from A and B is given by
∆xB=BP0-AP0=λ/3
⇒√(D²+d)²-D=λ/3
⇒D²+d²= D²+λ²/9 +2λD/3
We will neglect the term λ²/9, as it has a very small value.
∴d=2λD/3
(b) To calculate the intensity at P0,
consider the interference of light waves coming from all the three slits.
Path difference of the wave fronts reaching from A and C is given by
CP0-AP0=√D²+4d²-D
=√D²+8λD/3 - D
Using the value of d from part a
=D[ 1+8λ/3Dx2] -D
Expanding the value using binomial theorem and neglecting the higher order terms, we get:
D[1+8λ/3Dx 2 ..... } -D =4λ/3
So, the corresponding phase difference between the wave fronts from A and C is given by
ϕc=2πx/λ
=2π×4λ/3λ
⇒ϕc=8π/3 or 2π+2π/3
⇒ϕc=2π/3 …(1)
Again, ϕB=2πx/3λ
⇒ϕB=2πλ/3λ=2π3 …(2)
So, it can be said that light from B and C are in the same phase, as they have the same phase difference with respect to A.
Amplitude of wave reaching P0 is given by
A=√2a²+a²+2 x 2a×acos2π/3
=√4a²+a²-2a²
=√3
∴lpo- K (√3 r)²=√3 Kr²=√3
lHere, I is the intensity due to the individual slits and Ipo is the total intensity at P0.
Thus, the resulting amplitude is three times the intensity due to the individual slits.
Explanation:
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