Respuesta :
Answer:
(a) λ = 0.496 um (b) S =2π Δ d sinθ/ λ (c) I =gI₀ (d) For the central diffraction peak, a total of 5 interference maxima are present or available.
Note: find an attached copy of a part of the solution to the given question below.
Explanation:
Solution
Recall that:
d = 6.6 um
λ₀ =d/10
λ₀ = 6.6 um
Now,
(a) We find the wavelength λ of the light in water.
Thus,
λ water = (λ₀ )/n
= 0.66/1.33
So,
λ water = λ = 0.496 um
(b) We find the phase difference between the waves from slit 1 and 2
Now,
if a <<d and a<<λ
Then the path difference between the rays will be
Δ S₂N = Δ d sinθ
Thus, the phase difference becomes,
S = 2π Δ/λ is S= 2π Δ d sinθ/ λ
(c) The next step is to derive an expression for the intensity I as function of O and other relevant parameters.
Now,
Let p be the point where these two rays interfere with each other.
Thus,
The electric field vector coming out from slot and and slot 2 is
E₁= E₀₁ cos (ks₁ p - wt) i
E₂ = E₀₂ cos (ks₂ p - wt) i
Note: Kindly find an attached copy of a part of the solution to the given question below.
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![Ver imagen ogbe2k3](https://us-static.z-dn.net/files/dc9/ad0c47e026a84975e9e1e890fcbadef5.png)
![Ver imagen ogbe2k3](https://us-static.z-dn.net/files/d08/e43082326e928866276e8ea17de3bf4b.png)
![Ver imagen ogbe2k3](https://us-static.z-dn.net/files/d23/ba02689d15d4c3e3da2fda2f33534a84.png)
![Ver imagen ogbe2k3](https://us-static.z-dn.net/files/def/ac7e1eb39ad20f7ebe5015fe5171953e.png)