Question

In a YDSE, $d = 0.1mm$ and distance between slits and screen $D = 1m$ . At some point P on the screen, resulting intensity is equal to the intensity due to either of the individual slits. Then distance of P from the central maxima: $(\lambda = 6000{\rm A})$
A. 1mm
B. $0.5mm$
C. $0.25mm$
D. $2mm$

Answer 1

Hint We know that the intensity at any point in YDSE is given by $I = {I_{\max }}{\cos ^2}\dfrac{\phi }{2}$ where, $\phi $ is phase difference. Now using phase difference, we can calculate the path difference between the light waves. Once the path difference is known we can equate the equations of path difference to get the desired result.

Complete Step by step solution let the individual intensity of both the slits be ${I_o}$.
Now we know that the resultant intensity at any point in YDSE is given by,
$I = {I_{\max }}{\cos ^2}\dfrac{\phi }{2}......(1)$
Where, $\phi $ is phase difference.
If ${I_o}$ be the individual intensity of both the slits then,
$
 {I_{\max }} = {(\sqrt {{I_o}} + \sqrt {{I_o}} )^2} \\
 {I_{\max }} = {(2\sqrt {{I_o}} )^2} \\
 {I_{\max }} = 4{I_O}......(2) \\
 $
Now according to question intensity at point is given by intensity due to single slit i.e. $I = {I_o}$
From equation (1) and (2), we get
$
 {I_0} = 4{I_o}{\cos ^2}\dfrac{\phi }{2} \\
 {\cos ^2}\dfrac{\phi }{2} = \dfrac{1}{4} \\
 \cos \dfrac{\phi }{2} = \dfrac{1}{2} \\
    \\
 $
Hence from above we have,
$
  \dfrac{\phi }{2} = \dfrac{\pi }{3} \\
  \phi = \dfrac{{2\pi }}{3}......(3) \\
 $
Now, phase difference = $\dfrac{{2\pi }}{\lambda }$ path difference
          $\phi = \dfrac{{2\pi }}{\lambda }\Delta x$ where, $\Delta x$ is path difference.
Using equation (1) we get,
$
  \dfrac{{2\pi }}{3} = \dfrac{{2\pi }}{\lambda }\Delta x \\
  \therefore \Delta x = \dfrac{\lambda }{3}......\left( 4 \right) \\
 $
We know that the path difference is given by,
$\Delta x = \dfrac{{yd}}{D}......(5)$
Where,$y$ is the distance of point P from central maxima.
Hence finally using equation (4) and (5) we get,
$
 \dfrac{\lambda }{3} = \dfrac{{yd}}{D} \\
 y = \dfrac{{\lambda D}}{{3d}} \\
 $
On putting the values, we get
$
 y = \dfrac{{6000 \times {{10}^{ - 10}} \times 1}}{{3 \times 0.1 \times {{10}^{ - 3}}}} \\
 y = 2 \times {10^{ - 3}} \\
 y = 2mm \\
 $
Hence distance of point P from central maxima is $2mm$.

Option (C) is correct.

Note
${I_{\max }}$ and ${I_{\min }}$ is calculated by using the formula of resultant intensity at any point that is given by:
$I = {I_A} + {I_B} + 2{I_A}{I_B}\cos \delta $
For maximum intensity, $\delta = 1$
$
  \therefore I = {I_A} + {I_B} + 2{I_A}{I_B} \\
 I = {(\sqrt {{I_A}} + \sqrt {{I_B}} )^2} \\
 $

For minimum intensity, $\delta = - 1$
$
 \therefore I = {I_A} + {I_B} - 2{I_A}{I_B} \\
 I = {(\sqrt {{I_A}} - \sqrt {{I_B}} )^2} \\
 $