In a Young's double slit experiment, the intensities at two points, for the path difference (λ)/(4) and (λ)/(3) (λ being the wavelength of light used) are I_1 and I_2 respectively. If I_0 denotes the intensity produced

Correct answer: A. 3 Resultant intensity in Young's double slit experiment I=4 I _0 ^2 ((Δ φ)/(2)) For path difference (λ)/(4) phase difference, Δ φ=(2 π)/(λ) × (λ)/(4)=(π)/(4) I_ 1 =4 I _ 0 ^2((π)/(4))=2 I _ 0 For path difference (λ)/(3) I_2=4 I_0 ^2((2 π)/(λ) × (λ)/(3))=I_0 I _1+ I _2 I _0 =3

BITSAT 2023Paper 1

Exam: BITSAT
Year: 2023
Paper: Paper 1
Subject: Physics
Chapter: Wave Optics

BITSAT 2023 · Paper 1 · Physics · Wave Optics

In a Young's double slit experiment, the intensities at two points,...

In a Young's double slit experiment, the intensities at two points, for the path difference

\frac{\lambda}{4}

and

\frac{\lambda}{3}

(\lambda

being the wavelength of light used) are

I_1

and

I_2

respectively. If

I_0

denotes the intensity produced by each one of the individual slits, then \frac{

\mathrm{I}

_1+

\mathrm{I}

_2}{

\mathrm{I}

_0}=\ldots \ldots

Options

Correct answer

A. 3

Step-by-step solution

Resultant intensity in Young's double slit experiment

\mathrm{I}=4

\mathrm{I}

_0 \cos ^2 \left(\frac{\Delta \phi}{2}\right)<br />For path difference \frac{\lambda}{4} phase difference,<br />\Delta \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{4}<br />

\therefore

\mathrm{I}_{1}=4

\mathrm{I}

}_{0} \cos ^2\left(\frac{\pi}{4}\right)=2

\mathrm{I}

_{0}

For path difference

\frac{\lambda}{3}

I_2=4 I_0 \cos ^2\left(\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}\right)=I_0

\therefore \frac{

\mathrm{I}

_1+

\mathrm{I}

_2}{

\mathrm{I}

_0}=3

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