Consider the interference/diffraction pattern from a double-slit arrangement of slit separation d = 6.60 um and slit width a. The wavelength of the monochromatic light incident normally upon the slits is 2n = d/10 (in air). There is a filter (of negligible thickness) placed on slit 2, so that the magnitude of the EM wave emitted from it is half of that emitted from slit 1. The space between the slits and the screen is filled with water, whose index of refraction is n = 1.33 (you can take noir as 1.00). (a) What is the wavelength 2 of the light in water?
(b) If a << 1, What is the phase difference between the waves from slits 1 and 2?
(c) For a << 2, Derive an expression for the intensity / as a function of O and other relevant parameters, including the intensity at the center of the screen (where 0 = 0).
(d) Now suppose a = d/3. Redo part (b) above. How many interference maxima are present within the central diffraction peak? (Do not count the "clipped" maxima, if any.) (4) — E -2 d E ) в /Б/ = 1/5 | TT

(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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