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The width of one of the two slits in Young's double slit experiment is d while that of the other slit is xd. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9 : 4 then what is the value of x?
(Assume that the field strength varies according to the slit width.)

Asked in JEE Main 23rd Jan 2nd Shift 2025 · Intensity and coherent sources

Answer: (2) 5

Step-by-step solution

Field amplitude is proportional to width: (E₂)/(E₁) = x

(Iₘₐₓ)/(Iₘᵢₙ) = ((x+1)/(x-1))² = 9/4

(x+1)/(x-1) = 3/2, so 2x + 2 = 3x - 3

x = 5

Why the other options are wrong

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