Question

In a young’s double slit experiment,$d = 1mm,\lambda = 6000\mathop {\text{A}}\limits^{\text{0}}$ and$D = 1m$ (where$d,\lambda {\text{ }}and{\text{ }}D$have usual meaning). Each slit individually produces the same intensity on the screen. The minimum distance between two points on the screen having $75\%$ intensity of the maximum intensity is:
A) $0.45mm$
B) $0.40mm$
C) $0.30mm$
D) $0.20mm$

Answer 1

We know that intensity is directly proportional to square of the amplitude. Amplitude is the max displacement of the particle from its mean position. fringe width depends on the wavelength of light used, distance between the slits and distance between the screen and slit.

Complete step by step answer:
Given: Distance between the slits, $d = 1mm = 1 \times {10^{ - 3}}m$
Distance between screen and the slits,$D = 1m$
Wavelength of light used, $\lambda = 6000\mathop {\text{A}}\limits^{\text{0}}$
First of all we need to know the phase difference between two points on the screen where intensity reduces to $75\%$ of maximum intensity.
Intensity at any point is always directly proportional to the square of the amplitude. That is, $I \propto {a^2}$ the amplitude of the resultant wave , $R = 2a\cos \dfrac{\phi }{2}$
Therefore, resultant intensity of light at the given point due to both slits is given by,
$I \propto {R^2}$
Then
$\Rightarrow I \propto 4{a^2}{\cos ^2}\dfrac{\phi }{2}$
$\Rightarrow I = {I_{\max }}{\cos ^2}\left( {\dfrac{\phi }{2}} \right)$
We can now substitute the values of $I$ and ${I_{\max }}$ we get,
$\Rightarrow 75 = 100{\cos ^2}\left( {\dfrac{\phi }{2}} \right)$
simplifying the above equation we get,
$\Rightarrow {\cos ^2}\left( {\dfrac{\phi }{2}} \right) = \dfrac{3}{4}$
To remove the square in the left hand side we are taking square root on the right hand side, we get,
$\Rightarrow \cos \left( {\dfrac{\phi }{2}} \right) = \dfrac{{\sqrt 3 }}{2}$
After solving the above equation, we get
$\Rightarrow \dfrac{\phi }{2} = \dfrac{\pi }{6}$
$\Rightarrow \phi = \dfrac{\pi }{3}or{60^0}$
If phase difference is$\dfrac{\pi }{3}$, path difference would be,
Now we will calculate path difference between the waves,
$\Rightarrow \Delta = \dfrac{\lambda }{{2\pi }}\phi$
Substitute the value of phase difference in the above equation,
$\Rightarrow \Delta = \dfrac{\lambda }{{2\pi }} \times \dfrac{\pi }{3}$
With the help of the multiplication we can simplify the given equation,
$\Rightarrow \Delta = \dfrac{\lambda }{6}$
We know that,
$\Rightarrow x = \dfrac{D}{d}\Delta$
Substitute the given values we get,
$\Rightarrow x = \dfrac{1}{{1 \times {{10}^{ - 3}}}} \times \dfrac{\lambda }{6}$
Then,$x = {10^{ - 4}}m$
But, now we need to find calculate the minimum distance between two points on the screen having 75% intensity of the maximum intensity is that is,$AB = 2x = 2 \times {10^{ - 4}}$
$\therefore AB = 0.20mm$

Thus, the correct option is (D).

Additional information:
-Two sources must be coherent sources means two sources must emit waves of same frequency or wavelength having zero phase difference.
-The space between any two consecutive bright or dark fringes is called fringe width.
-From young’ double slit experiment, we can conclude that all bright and dark fringes are equally spaced

Note:
Condition for constructive interference (bright fringe): When crest of one superimpose with crest of another or trough of one wave superimpose with trough of another then maximum amplitude is formed and there is a creation of bright fringe is called constructive interference.
Phase difference $\phi = 2n\pi$ where n=0,1,2,3…..
Path difference, $\Delta = n\lambda$
Destructive interference (dark fringe): When crest of one wave superimposed with trough of another wave then resultant amplitude is minimum, and there is dark fringe is formed. This is called Destructive interference.
Phase difference $\phi = (2n - 1)\pi$ where n=0,1,2,3…..
Path difference, $\Delta = (2n - 1)\dfrac{\lambda }{2}$