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  •  In Young's double slit experiment the light emited from source has <img alt="I = 1.3 \times 10^{-6}" src="https://learn.careers360.com/latex-image/?I%20%3D%201.3%20%5Ctimes%2010%5E%7B-6%7

 In Young's double slit experiment the light emited from source has I = 1.3 \times 10^{-6}  m and the distance between the two slits is 2 mm. Distance between the screen and slit is 1 metre. Distance between third dark and fifth bright fringe will be:

Option: 1

3.2 mm


Option: 2

 1.63 mm


Option: 3

0.585 mm 


Option: 4

2.31 mm


Answers (1)

best_answer
 

 

.For Bright Fringes - 

By the principle of interference, condition for constructive interference is the path difference = nλ

\begin{array}{l}{\frac{xd}{D}=n \lambda} \\ \\ {\text { Here, } n=0,1,2 \ldots \ldots \text { indicate the order of bright fringes }} \\ {\text { So, } x=(\frac{n \lambda D}{d})} \\ \\ {\text { This equation gives the distance of the } n^{\text {th }} \text { bright fringe from the point O. }}\end{array}

 

For Dark fringes -

By the principle of interference, condition for destructive interference is the path difference =   \frac{(2n-1) \lambda}{2}

Here, n = 1,2,3 … indicate the order of the dark fringes.

                                   So, 

                                                                 x = \frac{(2n-1) \lambda D}{2d}

The above equation give the distance of the nth dark fringe from the point O. 

So, we can say that the alternate dark and bright fringe will be obtained on either side of the central bright fringe. 

 

    \begin{array}{l}{x _5=n \frac{\lambda D}{d}=\frac{5 \times 1.3 \times 10^{-6} \times 1}{2\times10^{-3}}=32.5 \times 10^{-4} \mathrm{m}} \\ \\ {x _3=(2 n-1) \frac{\lambda}{2} \frac{D}{d}=\frac{5 \times 1.3 \times 10^{-6} \times 1}{2 \times 2 \times 10^{-3}}=16.25 \times 10^{-4} \mathrm{m}} \\ \\ {\therefore x _5-\mathrm{x}_3 \approx 1.63 \mathrm{mm} .}\end{array}

The correct option is 2.

Posted by

Rishabh

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