Question

Intensities of light coming from the slit are I and 4I. Resultant intensity at point P with angular position theta=sin inverse Lawanda/3d

Answer 1

Answer:

Net intensity at point P is 3I

Explanation:

As we know that the angular position is given as

$\theta = sin^{-1}(\frac{\lambda}{3d})$

now we know that path difference at the given angular position is

$\Delta x = d sin\theta$

$\Delta x = d sin\theta = \frac{\lambda}{3}$

now phase difference at that position is given as

$\Delta \phi = \frac{2\pi}{\lambda}(\Delta x)$

$\Delta \phi = \frac{2\pi}{\lambda}(\frac{\lambda}{3})$

$\Delta \phi = \frac{2\pi}{3}$

now intensity at this position is given as

$I = I_1 + I_2 + 2\sqrt{I_1I_2} cos\phi$

$I_{net} = I + 4I + 4I cos(\frac{2\pi}{3})$

$I_{net} = 3I$

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Topic : Interference of light

https://brainly.in/question/9943187

Answer 2

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