In a Young's double slit experiment with light of wavelength λ the separation of slits is d and distance of screen is D such that D d λ. If the Fringe width is β, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is :

Question

In a Young's double slit experiment with light of wavelength λ the separation of slits is d and distance of screen is D such that D > > d > > λ. If the Fringe width is &beta;, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is : (A) (β/2) (B) (β/4) (C) (β/3) (D) (β/6)

Tardigrade Admin · Accepted Answer

For two identical slits, we can write the intensity as follows:

I = 4 I_{0} cos^{2} (\frac{ϕ}{2})

I_{ma x} = 4 I_{0}

Therefore, now, the half of maximum intensity is

\frac{I _{m a x}}{2} = 2 I_{0} = 4 I_{0} cos^{2} (\frac{ϕ}{2})

\Rightarrow cos (\frac{ϕ}{2}) = \frac{1}{2}

Therefore,

\frac{ϕ}{2} = \frac{π}{2}

\Rightarrow ϕ = \frac{π}{2}

That is,

\frac{2 π}{λ} Δ x = \frac{π}{2}

\Rightarrow Δ x = \frac{π}{4}

Now

y \frac{d}{D} = \frac{λ}{4}

\Rightarrow y = \frac{λ D}{4 d} = \frac{β}{4}

$\frac{β}{2}$

$\frac{β}{4}$

$\frac{β}{3}$

$\frac{β}{6}$

Q. In a Young's double slit experiment with light of wavelength λ the separation of slits is d and distance of screen is D such that D>>d>>λ. If the Fringe width is β, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is :

2β​

4β​

3β​

6β​

$\frac{β}{2}$

$\frac{β}{4}$

$\frac{β}{3}$

$\frac{β}{6}$