In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other - Physics MCQ AIEEE 2012 2026 | ExamTest.live

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In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If I_{m} be the maximum intensity, the resultant intensity I when they interfere at phase difference ϕ is given by:

Solution & Step-by-step Explanation

Let the amplitude of light from the first slit be A_{1} = a .Then amplitude from the second slit is A_{2} = 2 a .Maximum amplitude A_{m a x} = A_{1} + A_{2} = 3 a .Maximum intensity I_{m} \propto (3 a)^{2} = 9 a^{2} .Resultant intensity at phase difference ϕ : I = I_{1} + I_{2} + 2 I_{1} I_{2} cos ϕ I \propto a^{2} + (2 a)^{2} + 2 (a) (2 a) cos ϕ I = k (5 a^{2} + 4 a^{2} cos ϕ) = k a^{2} (5 + 4 cos ϕ) Since I_{m} = k (9 a^{2}) ⟹ k a^{2} = \frac{I _{m}}{9} . I = \frac{I _{m}}{9} (5 + 4 cos ϕ) Using cos ϕ = 2 cos^{2} \frac{ϕ}{2} - 1 : I = \frac{I _{m}}{9} (5 + 4 (2 cos^{2} \frac{ϕ}{2} - 1)) I = \frac{I _{m}}{9} (5 + 8 cos^{2} \frac{ϕ}{2} - 4) = \frac{I _{m}}{9} (1 + 8 cos^{2} \frac{ϕ}{2}) .

Try it yourself before checking the explanation above.

In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If I_{m} be the maximum intensity, the resultant intensity I when they interfere at phase difference ϕ is given by:

\frac{I_m}{9} (4 + 5 \cos \phi)

\frac{I_m}{3} (1 + 2 \cos^2 \frac{\phi}{2})

\frac{I_m}{5} (1 + 4 \cos^2 \frac{\phi}{2})

\frac{I_m}{9} (1 + 8 \cos^2 \frac{\phi}{2})

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