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 $\mathrm{I}_{1}$ and $\mathrm{I}_{2}$ respectively. If $\mathrm{I}_{0}$ denotes the intensity produced by each one of the individual slits, then $\frac{\mathrm{I}_{1}+\mathrm{I}_{2}}{\mathrm{I}_{0}}=$ \_\_\_\_

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Wave Optics

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

\frac{λ}{4}

and

\frac{λ}{3}

(

λ

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{I _{1} + I _{2}}{I _{0}} =

____

Answer: 3

Solution

The intensity at a point of phase difference

ϕ

in the double-slit pattern is

I = 4 I_{0} cos^{2} \frac{ϕ}{2}

, where

ϕ = \frac{2 π}{λ} \times (path difference)

. A path difference

\frac{λ}{4}

gives

ϕ = \frac{π}{2}

and

I_{1} = 4 I_{0} cos^{2} \frac{π}{4} = 2 I_{0}

, while

\frac{λ}{3}

gives

ϕ = \frac{2 π}{3}

and

I_{2} = 4 I_{0} cos^{2} \frac{π}{3} = I_{0}

. Adding these,

\frac{I _{1} + I _{2}}{I _{0}} = \frac{2 I _{0} + I _{0}}{I _{0}} = 3

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Integer answer

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