In the Young's double slit experiment, the intensity of light at a point on the screen is 'K', λ being the wavelength of light used. The intensity at a point where path difference is 6

Question

In the Young's double slit experiment, the intensity of light at a point on the screen is 'K', λ being the wavelength of light used. The intensity at a point where path difference is 6 λ ​ will be [ cos ( 6 π ​ ) = 2 3 ​ ​ ]

TestHub · Accepted Answer

The maximum intensity in Young's double slit experiment is given as $I_{ma x} = K$ .

The phase difference $ϕ$ is related to the path difference $Δ x$ by the formula:

$ϕ = \frac{2 π}{λ} Δ x$

Given the path difference $Δ x = \frac{λ}{6}$ , the phase difference is:

$ϕ = \frac{2 π}{λ} \times \frac{λ}{6} = \frac{π}{3}$

The intensity $I$ at any point on the screen is given by:

$I = I_{ma x} cos^{2} (\frac{ϕ}{2})$

Substituting the values of $I_{ma x}$ and $ϕ$ :

$I = K cos^{2} (\frac{π}{6})$

Since $cos (\frac{π}{6}) = \frac{3}{2}$ , we get:

$I = K (\frac{3}{2})^{2} = \frac{3 K}{4}$

Answer: $\frac{3 K}{4}$

Physics - Wave Optics Question with Solution | TestHub

Options:

Doubts & Discussion