Wave Optics

Question

In Young's double-slit experiment $\frac{d}{D} = 10^{- 4}$ (d = distance between slits, D = distance of screen from slits). At a point P on the screen resulting intensity is equal to the intensity due to individual slit I₀. Find the distance of point P from the central maximum. (λ = 6000Å).

Solution

Given,
Ratio of slit width to distance between the screen and slit,

\frac{d}{D} = 10^{- 4}

Wavelength of the light source,

λ

= 6000

\overset{o}{A}

and also,
Resultant intensity is equal to the intensity due to individual slit I_o.

This implies,

I = 4 I_{0} \cos^{2} (ϕ / 2)

i.e.,

I_{0} = 4 I_{0} \cos^{2} (ϕ / 2)

∴

\cos (ϕ / 2) = 1 / 2

\Rightarrow

ϕ / 2 = π / 3

or,

ϕ = \frac{2 π}{3}

and,
Relation between phase difference and path difference is,

ϕ = (\frac{2 π}{λ}) . ∆ x

or,

\frac{2 π}{3} = \frac{2 π}{λ} (\frac{yd}{D})

(∵ ∆ x = \frac{yd}{D})

∴

Y = \frac{λ}{3 \times d / D}

\Rightarrow

γ = \frac{6 \times 10^{- 7}}{3 \times 10^{- 4}} = 2 \times 10^{- 3} m

γ = 2 mm .

Therefore, the distance of point P (on the screen) from the central maximum is 2 mm .

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