In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is ?

Solution

Intensity of light = K units
Phase difference corresponding to λ is 2π and phase difference corresponding to λ/3 is

2 π / 3 .

Now, using the formula of intensity,

I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ

Let,

I_{1} = I_{2} = I_{0}

In the first case, when path difference is

λ

K = I_{0} + I_{0} + 2 I_{0} \cos 2 π = 4 I_{0}

In the second case, when path difference is

\frac{λ}{3}

K' = I_{0} + I_{0} + 2 I_{0} \cos \frac{2 π}{3}

= I_{0} + I_{0} - 2 I_{0} (\frac{1}{2}) = I_{0}

Now,

\frac{K'}{K} = \frac{I_{0}}{4 I_{0}} = \frac{1}{4}

K' = \frac{K}{4}

Therefore, intensity is one-fourth of it's initial value when path difference is λ/3.
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