Question

A parallel beam of visible light consisting of wavelengths ${\lambda }_1$ & ${\lambda }_2$ is incident on a standard YDSE apparatus with $d=1\text{mm},D=1\text{m}.$ $P$ is a point on the screen at a distance $y$ from center of screen $O.y=y_1$ is the nearest point above $O$ where the two maxima coincide and $y=y_2$ is the nearest point above $O$ where the two minima coincide. ${\beta }_1$ & ${\beta }_2$ are fringe width corresponding to wave length ${\lambda }_1$ & ${\lambda }_2.$

$\begin{array}{cc}\text{Column\_I}& \text{Column-II}\\ \text{(A)}{\beta }_1=0.3\text{mm},{\beta }_2=0.5\text{mm}& \text{(P) The}{2}^{nd}\text{nearest point above}O\text{where two maxima coincide is}y=2y_1\\ \text{(B)}{\beta }_1=0.3\text{mm},{\beta }_2=0.4\text{mm}& \text{(Q)}{y}_2\text{has no finite value}\\ \text{(C)}{\beta }_1=0.2\text{mm},{\beta }_2=0.4\text{mm}& \text{(R) The}{2}^{nd}\text{nearest point above}O\text{where the two minima coincide is}y=3y_2\\ \text{(D)}{\beta }_1=0.2\text{mm},{\beta }_2=0.6\text{mm}& \text{(S)}{y}_1=\text{LCM of}{\beta }_1\text{and}{\beta }_2\\ & \text{(T) The}{2}^{nd}\text{nearest point above}O\text{where two minima coincide is}y=2y_2\end{array}$
Which of the following option has the correct combination considering column-I and column-II ?

• A. $C\to R,T$
• B. $A\to Q,S$
• C. $D\to P,R,S$
• D. $B\to Q,T,S$

Answer 1

The correct option is C $D\to P,R,S$

For maxima the distance of $n^{th}$ bright fringe from center $O$ is
$y_n=\frac{nD\lambda }{d}=n\beta$
Given that $y=y_1$ is the nearest point above $O$, where $n_1^{th}$ maxima of wavelength ${\lambda }_1$ coincides with $n_2^{th}$ maxima of wavelength ${\lambda }_2$
$y_1=n_1{\beta }_1=n_2{\beta }_2=$ LCM of ${\beta }_1$ and ${\beta }_2$
For $y=2y_1=2n_1{\beta }_1=2n_2{\beta }_2$
Hence at this point both maxima again coincide.
Similary for minima, distance of $n^{th}$ dark fringe from center $O$ is
$y_n=\frac{(2n\pm 1)D\lambda }{2d}=(2n\pm 1)\frac{\beta }{2}$

Given that $y=y_2$ is the nearest point above $O$, where $n_1^{th}$ minima of wavelength ${\lambda }_1$ coincides with $n_2^{th}$ minima of wavelength ${\lambda }_2$
$y_2=(n_1-\frac{1}{2}){\beta }_1=(n_2-\frac{1}{2}){\beta }_2$
$\frac{{\beta }_1}{{\beta }_2}=\frac{n_2-\frac{1}{2}}{n_1-\frac{1}{2}}$
$\Rightarrow \frac{{\beta }_1}{{\beta }_2}=\frac{2n_2-1}{2n_1-1}$
Which will have a solution only if $\frac{{\beta }_1}{{\beta }_2}$ expressed as a proper fraction of form $\frac{\text{odd}}{\text{odd}}$.
Option $B\&C$ $:\frac{{\beta }_1}{{\beta }_2}$ is not of form $\frac{\text{odd}}{\text{odd}}$. hence not real value of $y_2$ exists.
Hence at some finite $y_2$ the two minima will coincide.
At $y=2y_2$ the two maxima (and not minima) will coincide.
$\therefore y=3y_2$ is the next nearest point where minima coincide.


$A\to P,R,S;:B\to P,Q,R,S:C\to P,Q,R,S:D\to P,R,S$