Question

Consider the situation shown in the figure. The two slits S₁ and S₂ placed symmetrically around the central line are illuminated by a monochromatic light of wavelength λ. The separation between the slits is d. The light transmitted by the slits falls on a screen ∑₁placed at a distance D from the slits. The slit S₃ is at the central line and the slit S₄ is at a distance z from S₃. Another screen ∑₂ is placed a further distance D away from ∑_1.Find the ratio of the maximum to minimum intensity observed on ∑₂ if z is equal to

(a) \[z = \frac{\lambda D}{2d}\]

(b) \[\frac{\lambda D}{d}\]

(c) \[\frac{\lambda D}{4d}\]

Answer 1

Given:
Separation between the two slits = d
Wavelength of the light = \[\lambda\]

Distance of the screen = D

The fringe width (β) is given by \[\beta = \frac{\lambda D}{d}\]

At S₃, the path difference is zero. So, the maximum intensity occurs at amplitude = 2a.

(a) When \[z = \frac{D\lambda}{2d}\]

The first minima occurs at S₄, as shown in figure (a).

With amplitude = 0 on screen ∑_2, we get

\[\frac{l_{max}}{l_{min}} = \frac{\left( 2a + 0 \right)^2}{\left( 2a - 0 \right)^2} = 1\]

(b) When \[z = \frac{D\lambda}{d}\]

The first maxima occurs at S₄, as shown in the figure.

With amplitude = 2a on screen ∑₂, we get

\[\frac{l_\max}{l_\min} = \frac{\left( 2a + 2a \right)^2}{\left( 2a - 2a \right)^2} =  \infty \]

(c) When \[z = \frac{D\lambda}{4d}\]

The slit S₄ falls at the mid-point of the central maxima and the first minima, as shown in the figure.

Intensity \[= \frac{l_\max}{2}\]

\[ \Rightarrow \text{Amplitude }= \sqrt{2}a\]

\[\therefore \frac{l_\max}{l_\min} = \frac{\left( 2a + \sqrt{2}a \right)^2}{\left( 2a - \sqrt{2}a \right)^2} = 34\]