Question

A monochromatic light source of wavelength is placed at S. Three slits S1,S2 and S3 are equidistant from the source S and the point P on the screen. S1p-S2P=\6 and S1p-S3p=2/3.If I be the intensity at P when only one slit is open , the intensity at P when all the three slits are open is

Answer 1

The intensity at P when all the three slits are open is I R = 3I

Explanation:

d(12) = 2π / λ × λ6 × 2 = 2π / 3

d(23) = 2πλ × λ2 × 2 = 2π

A' = √(√3 / 2)^2 + (3A /2 )^2 = √3 A

As we know that: I ∝ A^2

⇒ A^2R = 3A^2

⇒ I R = 3I

Hence the intensity at P when all the three slits are open is I R = 3I

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In a double slit experiment, when light of  wavelength 400 nm was used, the angular  width of the first minima formed on a screen  placed 1 m away, was found to be 0.2°. What  will be the angular width of the first minima, if  the entire experimental apparatus is immersed  in water? (μ(water) = 4/3)?

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