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# 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) $0.20mm$

Last updated date: 19th Sep 2024
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Hint: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.

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).

Phase difference $\phi = 2n\pi$ where n=0,1,2,3…..
Path difference, $\Delta = n\lambda$
Phase difference $\phi = (2n - 1)\pi$ where n=0,1,2,3…..
Path difference, $\Delta = (2n - 1)\dfrac{\lambda }{2}$