Question

Consider the situation shown in figure (17-E6). The two slits $S_1$ and $S_2$ placed symmetrically around the central line are illuminated by a monochromatic light of wavelength $\lambda$ the separation between the slits is d. The light transmitted by the slits falls on a screen ${\sum }_1$ placed at a distance D from the slits. The slit$S_3$ is at the central line and the slit $S_4$ is at a distane z rom $S_3$ Another screen ${\sum }_2$ is palced a further distance D away from ${\sum }_1$ find the ration of thmaximum to minimum intensity obsered on ${\sum }_2$ if z is equal to
(a) $z=\frac{\lambda D}{2d}$
(b) $\frac{\lambda }{d},$
(c) $\frac{\lambda D}{4d}.$

Answer 1

(i) When $z=\frac{D\lambda }{2d},at{S}_4$
Minimum ntensity occurs, (dark fringe)
$\Rightarrow Aplitude=0$

At $S_3$ pat difference = 0
$\Rightarrow$ Maximum intensity occurs.
$\Rightarrow$Amplitude = 2r
So, on ${\sum }^2$ screen,
$\frac{l_{max}}{l_{min}}=\frac{(2r+0{)}^2}{(2r-0{)}^2}=1$
(ii) When, $z=\frac{\lambda D}{d}$
At $S_4$ maximum ntensity occurs.
$\Rightarrow Amplitude=\sqrt{2}r$
At $S_3$ also maximum ntensity occurs.
$\frac{l_{max}}{l_{min}}=\frac{(2r+2r{)}^2}{(2r-2r{)}^2}=\infty \left(iii\right)when,Z=\frac{\lambda D}{4d},At{S}_4$
Intensity $=\frac{l_{max}}{2}$
$\Rightarrow$ Amplitude $=\sqrt{2r}$
$\therefore$ At $S_3$ Intesnsity is minimum.
$\Rightarrow$ Amplitude = 2r
$\frac{l_{max}}{l_{min}}=\frac{(2r+\sqrt{2r}{)}^2}{(2r-\sqrt{2r}{)}^2}=34$