Question

A monochromatic parallel beam of light of wavelength $\lambda$ is incident normally on the plane containing slits $S_1$ and $S_2$. The slits are of unequal width such that intensity only due to one slit on screen is four times that only due to the other slit. The screen is placed along y-axis as shown in figure. The distance between slits is $d$ and that between the screen and slits is $D$. Match the statements in column I with results in Column II

Answer 1

The maximum intensity of the interference pattern is $(\sqrt{{}_{4I_0}}+\sqrt{I_0}{)}^2=9I_0$

For A,

$I_0=I_0+4I_0+2\left(\sqrt{I_0}\sqrt{4I_0}cos\delta \right)$

$⟹cos\delta =\pi ,3\pi ,5\pi ,7\pi ....$

Path difference at the points of such intensity=$\delta \times \frac{\lambda }{2\pi }=\frac{\lambda }{2},\frac{3\lambda }{2},\frac{5\lambda }{2},\frac{7\lambda }{2}....$

Positions of such points=path difference $\times \frac{D}{d}=$

$=\frac{D\lambda }{2d},\frac{3D\lambda }{2d},\frac{5D\lambda }{2d},\frac{7D\lambda }{2d},....$

Hence, distance between such points=$\frac{D\lambda }{d},\frac{2D\lambda }{d},\frac{3D\lambda }{2d}$

For B,

$3I_0=I_0+4I_0+4I_{0}cos\delta$

$⟹\delta =\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3}....$

Similarly as done above,

Distance between points=$\frac{D\lambda }{3d}$,$\frac{D\lambda }{d},\frac{2D\lambda }{d},\frac{3D\lambda }{2d}$

For C,

$5I_0=I_0+4I_0+4I_{0}cos\delta$
$⟹\delta =\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2}...$

Distance between points is $\frac{D\lambda }{d},\frac{2D\lambda }{d},\frac{3D\lambda }{2d}$.

For D,

$7I_0=I_0+4I_0+4I_{0}cos\delta$

$⟹\delta =\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{11\pi }{3}..$

Distance between points is $\frac{D\lambda }{3d}$,$\frac{D\lambda }{d},\frac{2D\lambda }{d},\frac{3D\lambda }{2d}$