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

To solve this problem, we need to understand the effect of introducing a glass plate in one of the paths of the interfering beams in Young's Double Slit Experiment.

In Young's Double Slit Experiment, the condition for maxima (bright fringes) is given by the equation:

\(d \sin \theta = n \lambda\)

where \(d\) is the slit separation, \(\theta\) is the angle of diffraction, \(n\) is the order of the maximum, and \(\lambda\) is the wavelength of light.

When a glass plate of thickness \(t\) and refractive index \(\mu\) is introduced in the path of one of the beams, it introduces an additional phase difference. The optical path length in the medium of refractive index \(\mu\) is given by:

\(OP = \mu t\)

The introduction of the glass plate shifts the path difference by:

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

For the central maximum to remain unchanged, the path difference due to the glass plate should be an integral multiple of the wavelength:

 

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

Since the intensity at the central maximum remains unchanged, \(m = 1\). Therefore:

\((\mu - 1) t = \lambda\)

Given that the thickness of the glass plate is \(t = x \lambda\), substituting this into the equation, we have:

\((\mu - 1) x \lambda = \lambda\)

Simplifying this gives:

\(x = \frac{1}{\mu - 1}\)

For the given refractive index \(\mu = 1.5\),

\(x = \frac{1}{1.5 - 1} = \frac{1}{0.5} = 2\)

The value of \(x\) is therefore 2.