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. The distance between slits is $d$ and that between screen and slit is $D$. Match the statements in column-I with results in column-II.


$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \begin{array}{c}\text{(A)The distance between two points on screen having equal}\\ \text{intensities, such that intensity at those points is}(\frac{1}{9}{)}^{th}\text{of maximum intensity.}\end{array}& \text{(p)}\frac{D\lambda }{3d}\\ \begin{array}{c}\text{(B)The distance between two points on screen having equal}\\ \text{intensities, such that intensity at those points is}(\frac{3}{9}{)}^{th}\text{of maximum intensity.}\end{array}& \text{(q)}\frac{D\lambda }{d}\\ \begin{array}{c}\text{(C)The distance between two points on screen having equal}\\ \text{intensities, such that intensity at those points is}(\frac{5}{9}{)}^{th}\text{of maximum intensity.}\end{array}& \text{(r)}\frac{2\lambda D}{d}\\ \begin{array}{c}\text{(D)The distance between two points on screen having equal}\\ \text{intensities, such that intensity at those points is (}\frac{7}{9}{)}^{th}\text{of maximum intensity.}\end{array}& \text{(s)}\frac{3\lambda D}{d}\\ & \text{(t)}\frac{2\lambda D}{3d}\end{array}$

• A. $A–q,s,B–q,r,s,C–q,t,D–p,r,s$
• B. $A–q,r,s,B–p,q,r,s,tC–q,r,s,D–p,q,r,s,t$
• C. $A–q,s,B–q,r,s,C–q,r,tD–p,r,s$
• D. $A–p,s,t,B–p,r,s,C–q,s,D–p,r,s$

Answer 1

The correct option is B $A–q,r,s,B–p,q,r,s,tC–q,r,s,D–p,q,r,s,t$

Resultant intensity $I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$
Given $I_1=4I_2$
Let $I_2=I_0,I_1=4I_0$
$\Rightarrow I=4I_0+I_0+4I_0\cos \varphi =5I_0+4I_0\cos \varphi \Rightarrow I_{max}=9I_0$
$\text{(A)}I=\frac{1}{9}{I}_{max}\Rightarrow I_0=5I_0+4I_0\cos \varphi \Rightarrow \cos \varphi =-1\varphi =\pi ,3\pi ,5\pi ,....\text{Distance of point from central maxima on either sides}y=\frac{D}{d}\frac{\lambda }{2\pi }\Delta \varphi =\frac{\lambda D}{2d},\frac{3\lambda D}{2d},\frac{5\lambda D}{2d},\frac{7\lambda D}{2d}\text{The distance between two points having equal intensities, such that intensity at those points is}(\frac{1}{9}{)}^{th}\text{of maximum intensity.}\Rightarrow \Delta y=\frac{\lambda D}{d},\frac{2\lambda D}{d},\frac{3\lambda D}{d},\frac{4\lambda D}{d}....$
Hence $q,r,s$

$\text{(B)}I=\frac{3}{9}{I}_{max}\Rightarrow 3I_0=5I_0+4I_0\cos \varphi \Rightarrow \cos \varphi =-\frac{1}{2}\varphi =\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3},\frac{14\pi }{3},\frac{16\pi }{3}\text{Distance of point from central maxima on either sides}y=\frac{D}{d}\frac{\lambda }{2\pi }\Delta \varphi =\frac{\lambda D}{3d},\frac{2\lambda D}{3d},\frac{4\lambda D}{3d},\frac{5\lambda D}{3d},\frac{7\lambda D}{3d},\frac{8\lambda D}{3d}\text{The distance between two points having equal intensities, such that intensity at those points is}(\frac{3}{9}{)}^{th}\text{of maximum intensity.}\Rightarrow \Delta y=\frac{\lambda D}{3d},\frac{2\lambda D}{3d},\frac{\lambda D}{d},\frac{5\lambda D}{3d},\frac{2\lambda D}{d},\frac{8\lambda D}{3d},\frac{3\lambda D}{d},\frac{10\lambda D}{3d}$
Hence $p,q,r,s,t$

$\text{(C)}I=\frac{5I_{max}}{9}=5I_0\Rightarrow \cos \varphi =0\Rightarrow \theta =\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2},\frac{9\pi }{2},\frac{11\pi }{2}...\text{Distance of point from central maxima on either sides}y=\frac{D}{d}\frac{\lambda }{2\pi }\Delta \varphi =\frac{\lambda D}{4d},\frac{3\lambda D}{4d},\frac{5\lambda D}{4d},\frac{7\lambda D}{4d},\frac{9\lambda D}{4d},\frac{11\lambda D}{4d}\text{The distance between two points having equal intensities, such that intensity at those points is}(\frac{5}{9}{)}^{th}\text{of maximum intensity.}\Rightarrow \Delta y=\frac{\lambda D}{2d},\frac{\lambda D}{d},\frac{3\lambda D}{2d},\frac{2\lambda D}{d},\frac{5\lambda D}{2d},\frac{3\lambda D}{d}...$
Hence $q,r,s$

$\text{(D)}I=\frac{7}{9}{I}_{max}=7I_0\Rightarrow \cos \varphi =\frac{1}{2}\Rightarrow \varphi =\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{11\pi }{3},...\text{Distance of point from central maxima on either sides}y=\frac{D}{d}\frac{\lambda }{2\pi }\Delta \varphi =\frac{\lambda D}{6d},\frac{5\lambda D}{6d},\frac{7\lambda D}{6d},\frac{11\lambda D}{6d}...\text{The distance between two points having equal intensities, such that intensity at those points is}(\frac{7}{9}{)}^{th}\text{of maximum intensity.}\Delta y=\frac{\lambda D}{3d},\frac{2\lambda D}{3d},\frac{\lambda D}{d},\frac{4\lambda D}{3d},\frac{5\lambda D}{3d},\frac{2\lambda D}{d},\frac{7\lambda D}{3d},\frac{8\lambda D}{3d},\frac{3\lambda D}{d}$
Hence $p,q,r,s,t$