Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A Young's double-slit experiment is performed using monochromatic light of wavelength
. The intensity of light at a point on the screen, where the path difference is
, is
units. The intensity of light at a point where the path difference is
is given by
,$ where
is an integer. The value of
is
Text Solution
Verified by ExpertsThe correct answer is:
D
Step 1: In Young's double-slit experiment, the intensity of light at a point on the screen is given by the formula:
I = I_0 imes ext{cos}^2rac{eta}{2}
where β = rac{2 ext{π}}{ ext{λ}} imes d imes sin(θ),
and I_0 is the maximum intensity.
Step 2: Given the path difference Δx = rac{ ext{nλ}}{6}, we can set this into the above equation:
Δx = d imes sin(θ), hence:
sin(θ) = rac{Δx}{d} = rac{nλ/6}{d}.
Step 3: The intensity at the new path difference Δx = nλ/12 is equivalent to n = n/2.
Thus, Δx = nλ/6 leads to the calculation of n.
Step 4: Solving further gives n = 2. Hence, the value of n is 2.
Therefore, the correct value of n is D..
I = I_0 imes ext{cos}^2rac{eta}{2}
where β = rac{2 ext{π}}{ ext{λ}} imes d imes sin(θ),
and I_0 is the maximum intensity.
Step 2: Given the path difference Δx = rac{ ext{nλ}}{6}, we can set this into the above equation:
Δx = d imes sin(θ), hence:
sin(θ) = rac{Δx}{d} = rac{nλ/6}{d}.
Step 3: The intensity at the new path difference Δx = nλ/12 is equivalent to n = n/2.
Thus, Δx = nλ/6 leads to the calculation of n.
Step 4: Solving further gives n = 2. Hence, the value of n is 2.
Therefore, the correct value of n is D..
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