In Young’s double slit experiment performed using a monochromatic light of wavelength λ, when a glass plate (μ = 1.5) of thickness xλ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be:
- 3
- 2
- 1.5
- 0.5
The Correct Option is B
Approach Solution - 1
In Young's double slit experiment, when a glass plate is introduced in the path of one of the beams, it causes an additional path difference due to the change in optical path length. This can result in a shift of the interference pattern.
The formula for the optical path difference introduced by a glass plate is given by:
\(\Delta = (\mu - 1) t\)
where \(\mu\) is the refractive index of the glass plate and \(t\) is its thickness.
For the intensity at the original central maximum to remain unchanged, the introduced path difference \(\Delta\) must be an integral multiple of the wavelength \(\lambda\). Therefore:
\((\mu - 1) t = n\lambda\)
Given: \(\mu = 1.5\) and \(t = x\lambda\).
So, substituting these values, we have:
\(((1.5 - 1) x\lambda) = n\lambda\)
This simplifies to:
\(0.5x\lambda = \lambda\)
Cancelling \(\lambda\) from both sides, we get:
\(0.5x = 1\)
Thus, \(x = \frac{1}{0.5} = 2\)
Therefore, the value of \(x\) is 2, which corresponds to the condition given in the problem statement.
Approach Solution -2
The correct answer is (B) : 2
For the intensity to remain same the position must be of a maxima so path difference must be nλ so
(1.5 – 1)xλ = nλ
x = 2n (n = 0, 1, 2 …)
So, value of x will be
x = 0, 2, 4, 6…
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Concepts Used:
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- Thus, an interference pattern takes place when both the slits S1 and S2 are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged, to reach point P the light waves from slits S1 and S2 must travel at different distances. It implies that there is a path difference in the Young double-slit experiment between the two slits S1 and S2.
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