Question

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:

• A. 3
• B. 2
• C. 1.5
• D. 0.5

Answer 1

The Correct Option is B

 In the Young's double slit experiment, when a glass plate is introduced in the path of one of the interfering beams, it causes a change in the optical path length. This can potentially lead to a shift in the interference pattern. However, when the intensity at the position where the central maximum previously occurred remains unchanged, it means there is no shift in the central maximum.

To understand this scenario, we have to ensure that the phase difference introduced by the glass plate cancels out the shift due to its thickness.

The optical path difference introduced due to a glass plate of thickness \(t\) and refractive index \(\mu\) is given by:

\(\Delta = (\mu - 1)t\).

In the case of the central maximum position remaining unchanged, this optical path difference should be an integral multiple of the wavelength \(\lambda\). For the minimum thickness, using this condition, we can write:

\((\mu - 1)t = m\lambda,\) where \(m\) is an integer.

Given that \(\mu = 1.5\), substituting in the equation above gives:

\((1.5 - 1)t = m\lambda\)

\(0.5t = m\lambda\)

\(t = \frac{m\lambda}{0.5} = 2m\lambda\)

Thus, the thickness \(t\) of the glass plate is \(x\lambda = 2m\lambda\). In this scenario, the smallest non-zero integer value of \(m\) that ensures no shift of the central maximum is 1. Therefore, \(x = 2\).

Hence, the correct value of \(x\) is 2. Therefore, the correct answer is option "

2

".