In a Young's double slit experiment, the distance of the screen from the slits is 2.0 m , the separation between the slits is 2.0 mm , and the wavelength of the light used is 600 nm . Given that the intensity at the

Correct answer: A. 0.05 W m^ -2 Given D = 2.0 m , d = 2.0 × 10^ -3 m , λ = 600 × 10^ -9 m , I_0 = 0.20 W m^ -2 , and y = 0.5 × 10^ -2 m . The path difference at distance y from the central maximum is given by: Δ x = (y d)/(D) Substituting the values: Δ x = 0.5 × 10^ -2 × 2.0 × 10^ -3 2.0 = 5 × 10^ -6 m The corresponding phase difference is: Δ φ = (2π)/(λ) Δ x Δ φ = 2π 600 × 10^ -9 × 5 × 10^ -6 = (50π)/(3) The intensity at this point is given by: I = I_0 ^2 ((Δ φ)/(2)) I = 0.20 ^2 ((25π)/(3)) I = 0.20 ^2 (8π + (π)/(3)) = 0.20 ^2((π)/(3)) I = 0.20 × ((1)/(2))^2 = 0.05 W m^ -2

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Light Waves

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Light Waves

In a Young's double slit experiment, the distance of the screen...

In a Young's double slit experiment, the distance of the screen from the slits is

2.0 \text{ m}

, the separation between the slits is

2.0 \text{ mm}

, and the wavelength of the light used is

600 \text{ nm}

. Given that the intensity at the centre of the central maximum is

0.20 \text{ W m}^{-2}

, determine the intensity at a point located

0.5 \text{ cm}

away from this centre along the width of the fringes.

Options

A $0.05 \text{ W m}^{-2}$
B $0.15 \text{ W m}^{-2}$
C $0.10 \text{ W m}^{-2}$
D $0.20 \text{ W m}^{-2}$

Correct answer

A. $0.05 \text{ W m}^{-2}$

Step-by-step solution

Given

D = 2.0 \text{ m}

d = 2.0 \times 10^{-3} \text{ m}

\lambda = 600 \times 10^{-9} \text{ m}

I_0 = 0.20 \text{ W m}^{-2}

, and

y = 0.5 \times 10^{-2} \text{ m}

.

The path difference at distance

y

from the central maximum is given by:

\Delta x = \dfrac{y d}{D}

Substituting the values:

\Delta x = \dfrac{0.5 \times 10^{-2} \times 2.0 \times 10^{-3}}{2.0} = 5 \times 10^{-6} \text{ m}

The corresponding phase difference is:

\Delta \phi = \dfrac{2\pi}{\lambda} \Delta x

\Delta \phi = \dfrac{2\pi}{600 \times 10^{-9}} \times 5 \times 10^{-6} = \dfrac{50\pi}{3}

The intensity at this point is given by:

I = I_0 \cos^2 \left(\dfrac{\Delta \phi}{2}\right)<br><br>

I = 0.20 \cos^2 \left(\dfrac{25\pi}{3}\right)

<br><br>

I = 0.20 \cos^2 \left(8\pi +

\dfrac{\pi}{3}\right) =

0.20 \cos^2\left(\dfrac{\pi}{3}\right)

I = 0.20 \times \left(\dfrac{1}{2}\right)^2 = 0.05 \text{ W m}^{-2}

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