Answer

Verified

402.9k+ views

**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.

**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, $\Delta = (2n - 1)\dfrac{\lambda }{2}$

Recently Updated Pages

Draw a labelled diagram of DC motor class 10 physics CBSE

A rod flies with constant velocity past a mark which class 10 physics CBSE

Why are spaceships provided with heat shields class 10 physics CBSE

What is reflection Write the laws of reflection class 10 physics CBSE

What is the magnetic energy density in terms of standard class 10 physics CBSE

Write any two differences between a binocular and a class 10 physics CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Name 10 Living and Non living things class 9 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Write the 6 fundamental rights of India and explain in detail