In a Young's double slit experiment the intensity at a point where the path difference is λ/6 ( λ being the wavelength of the light used) is I . If I_ - Physics MCQ AIEEE 2007 2026 | ExamTest.live

mediumMCQAIEEE 20072026Physics

1 attempts0% success rate1 mark

In a Young's double slit experiment the intensity at a point where the path difference is λ /6 (λ being the wavelength of the light used) is I . If I_{0} denotes the maximum intensity, I / I_{0} is equal to:

Solution & Step-by-step Explanation

Phase difference (ϕ) is related to path difference (Δ x) by: ϕ = \frac{2 π}{λ} Δ x Given Δ x = λ /6 : ϕ = \frac{2 π}{λ} \cdot \frac{λ}{6} = \frac{π}{3} = 6 0^{\circ} The intensity I in interference is given by: I = I_{0} cos^{2} (\frac{ϕ}{2}) I = I_{0} cos^{2} (3 0^{\circ}) = I_{0} (\frac{3}{2})^{2} I = I_{0} (\frac{3}{4}) ⟹ \frac{I}{I _{0}} = \frac{3}{4}

Try it yourself before checking the explanation above.

In a Young's double slit experiment the intensity at a point where the path difference is λ /6 (λ being the wavelength of the light used) is I . If I_{0} denotes the maximum intensity, I / I_{0} is equal to:

1/2

3 /2

1/4

3/4

Practice with timed mock tests and track your performance across Physics.