Question

A parallel beam of light consisting of two wavelengths ${\lambda }_1=4000{\text{A}}^{\circ }$ and ${\lambda }_2=8000{\text{A}}^{\circ }$ is incident perpendicular to plane of both slits in a typical young's double slit experiment. The separation between both slits is $d=2\text{mm}$ and the distance between slits and screen is $D=1\text{m}$. In each situation of column-I a point ${}^{\prime }{\text{p}}^{\prime }$ on screen is specified by its distance ${}^{\prime }{l}^{\prime }$ from central bright on screen. Match the proper entries from column-II to column-I.
Column-I Column-II

$\left(\text{P}\right)$ At $\text{p}$ such that $l=0$.	$\left(1\right)$ Intensity is maximum for ${\lambda }_1=4000{\text{A}}^{\circ }$
$\left(\text{Q}\right)$ At $\text{p}$ such that $l=0.1\text{mm}$.	$\left(2\right)$ Intensity is minimum for ${\lambda }_1=4000{\text{A}}^{\circ }$
$\left(\text{R}\right)$ At $\text{p}$ such that $l=0.2\text{mm}$.	$\left(3\right)$ Intensity is maximum for ${\lambda }_2=8000{\text{A}}^{\circ }$
$\left(\text{S}\right)$ At $\text{p}$ such that $l=0.4\text{mm}$.	$\left(4\right)$ Intensity is minimum for ${\lambda }_2=8000{\text{A}}^{\circ }$

• A. $\text{P}\to (1,3);\text{Q}\to \left(2\right);\text{R}\to \left(4\right);\text{S}\to (1,3)$
• B. $\text{P}\to (1,3);\text{Q}\to \left(2\right);\text{R}\to \left(4\right);\text{S}\to \left(3\right)$
• C. $\text{P}\to \left(2\right);\text{Q}\to \left(4\right);\text{R}\to \left(4\right);\text{S}\to (1,3)$
• D. $\text{P}\to (1,3);\text{Q}\to \left(2\right);\text{R}\to \left(1\right);\text{S}\to (1,2)$

Answer 1

The correct option is A $\text{P}\to (1,3);\text{Q}\to \left(2\right);\text{R}\to \left(4\right);\text{S}\to (1,3)$


A minima corresponds to an absent wavelength.

So let us find Position order of minima using, $$y_n=\frac{(2n-1)\lambda D}{2d}$$From the data given in the question,

Case-I : $\lambda =4000{\text{A}}^{\circ }$

$n=\frac{1}{2}[\frac{2d{y}_n}{D\lambda }+1]=\frac{1}{2}[\frac{2\times 2\times {10}^{-3}\times y_n}{1\times 4\times {10}^{-7}}+1]$

For $y_n=0\to n=0.5$

$y_n=0.1\text{mm}\to n=1$

$y_n=0.2\text{mm}\to n=1.5$

$y_n=0.4\text{mm}\to n=2.5$

Case-II : $\lambda =8000{\text{A}}^{\circ }$

$n=\frac{1}{2}[\frac{2d{y}_n}{D\lambda }+1]=\frac{1}{2}[\frac{2\times 2\times {10}^{-3}\times y_n}{1\times 8\times {10}^{-7}}+1]$

For $y_n=0\to n=0.5$

$y_n=0.1\text{mm}\to n=0.75$

$y_n=0.2\text{mm}\to n=1$

$y_n=0.4\text{mm}\to n=1.5$

A maxima corresponds to presence of wavelength.

So, let us find position order of maxima using,$$y_n=\frac{n\lambda D}{d}$$ From the data given in the question,

Case-I : $\lambda =4000{\text{A}}^{\circ }$

$n=\left[\frac{d{y}_n}{D\lambda }\right]=\left[\frac{2\times {10}^{-3}\times y_n}{1\times 4\times {10}^{-7}}\right]$

For $y_n=0\to n=0$

$y_n=0.1\text{mm}\to n=0.5$

$y_n=0.2\text{mm}\to n=1$

$y_n=0.4\text{mm}\to n=2$

Case-II : $\lambda =8000{\text{A}}^{\circ }$

$n=\left[\frac{d{y}_n}{D\lambda }\right]=\left[\frac{2\times {10}^{-3}\times y_n}{1\times 8\times {10}^{-7}}\right]$

For $y_n=0\to n=0$

$y_n=0.1\text{mm}\to n=0.25$

$y_n=0.2\text{mm}\to n=0.5$

$y_n=0.4\text{mm}\to n=1$

Hence, option (a) is the correct answer.