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  • (a) Can two independent monochromatic light sources be used to obtain a steady interference pattern? Justify your answer. (b) In a Young's double-slit experiment, explain the formation of interference fringes and obtain an expression for

(a) Can two independent monochromatic light sources be used to obtain a steady interference pattern? Justify your answer.

(b) In a Young's double-slit experiment, explain the formation of interference fringes and obtain an expression for the fringe width.

(c) In an interference experiment using monochromatic light of wavelength \lambda, the intensity of light of a point, where the path difference is \lambda, on the screen is K units. Find out the intensity of light at a point when path difference is \frac{\lambda}{4}.

 

 

 

 
 

Answers (1)

(a) No, two independent monochromatic light source cannot be used to obtain a steady interference pattern cause:

Light waves coming from two independent sources will not have any fixed phase relationship.

Also, light waves coming from two independent sources will be incoherent. For interference to take place the sources must be coherent (i.e, having the same frequency and wavelength).

(b)

Light from a bright source was made to fall on two silts S1 and S2 made on an opaque screen through slit S. When light waves from two illuminated silts are induced on screen, the path travelled by each wave is different. Thus leads to a phase difference in the two light waves. The path difference is different for each point on the screen. As a result, the intensity is different at all points. This results in the formation of bright dark fringes on the screen.

When S1 and S2 are two coherent sources and for an arbitrary point P on the line GG' to correspond to the maximum we must have

S_{2}P-S_{1}P=n\lambda; n = 0,1,2...

Now 

\left ( S_{2}P \right )^{2}-\left ( S_{1}P \right )^{2}=\left [ D^{2}+\left ( x +\frac{d}{2} \right )^{2} \right ]-\left [ D^{2}+\left ( x -\frac{d}{2} \right )^{2} \right ]=2xdwhere S_{1}S_{2}=d and OP = x

S_{2}P-S_{1}P= \frac{2xd}{S_{2}P+S_{1}P}

If x, d <<D, then negligible error will be introduced if S_{2}P+S_{1}P is replaced by 2D.

\therefore S_{2}P-S_{1}P\cong \frac{xd}{D}

hence we will have constructive interference resulting in a bright region when

x = x_{n}= \frac{n\lambda D}{d}; n=0,\pm 1,\pm 2....                         .......(1)

We will have a dark region near

x = x_{n}= \left ( n+\frac{1}{2} \right )\frac{\lambda D}{d}; n=0,\pm 1,\pm 2....                .......(2)

Fringe width \beta = x_{n+1}-x_{n}

\therefore \beta = \frac{\lambda D}{d}

From equation (1) and (2) we understood that dark and bright fringe are equally spaced.

(c)  Phase \ difference=\frac{2\pi}{\lambda}\times path \ difference 

At path difference \lambda

\phi =\frac{2\pi}{\lambda}\times\lambda =2\pi

Intensity , I = 4I_{0}\cos ^{2}\frac{\phi}{2}

I = K (given)

K = 4I_{0}\cos ^{2}\frac{2\pi}{2}=4I_{0}

I_{0}=\frac{K}{4}

At path difference \frac{\lambda }{4}

Intensity , I' = 4I_{0}\cos ^{2}\frac{\phi}{2}

               I' = 4I_{0}\cos ^{2}\frac{1}{2}\times \frac{\pi }{2}

              I' = 4I_{0}\cos ^{2}\times \frac{\pi }{4}=4\times \frac{K}{4}\cos ^{2}\times \frac{\pi }{4}

            I' =\frac{K}{2}

Posted by

Safeer PP

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