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 ∑₁ placed 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) $z=\frac{\lambda \mathrm{D}}{2d}$
(b) $\frac{\lambda \mathrm{D}}{d}$
(c) $\frac{\lambda \mathrm{D}}{4d}$
Figure

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.
$$\begin{gathered} \mathrm{Intensity}=\frac{l_{\max }}{2} \\ \Rightarrow \mathrm{Amplitude}=\sqrt{2}a \end{gathered}$$
$\therefore \frac{l_{\max }}{l_{\min }}=\frac{{\left(2a+\sqrt{2}a\right)}^2}{{\left(2a-\sqrt{2}a\right)}^2}=34$