Question

In a Young's double slit experiment, the ratio of the slits width 4:1 . the ratio of the intensity of maxima to minima close to the central fringe on the screen , will be ​

Answer 1

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Answer 2

Given :

• the ratio of the slits width = 4:1

To find :

• the ratio of the intensity of maxima to minima close to the central fringe on the screen.

Solution :

Width ratio is given as = 4:1

Now , we know that Intensity is directly proportional to width.

Therefore , Ratio of intensity and width will be equal.

So, intensity ratio = 4:1

Now , let intensity from slit 1 and slit 2 be 4x and x.

Now , net intensity is :-

$\rm I_{net} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \: cos \phi....(1)$

Now , to get maximum intensity put cosΦ = 1 in (1)

$\rm \bf I_{max} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}}$

Now , put :-

• I_1 = 4x
• I_1 = x

$\rm \implies I_{max} = 4x + x+ 2 \sqrt{4x \times x}$

$\rm \implies I_{max} =5x+ 2 \sqrt{4 {x}^{2} }$

$\rm \implies I_{max} =5x+ 2.2x$

$\rm \implies I_{max} =5x+ 4x$

$\rm \implies I_{max} = 9x$

Now , to get minimum intensity put cosΦ = -1 in (1)

$\rm \bf I_{min} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$

Now , put :-

• I_1 = 4x
• I_1 = x

$\rm \implies I_{min} = 4x + x - 2 \sqrt{4x \times x}$

$\rm \implies I_{min} = 5x - 2 \sqrt{4 {x}^{2} }$

$\rm \implies I_{min} = 5x - 2 .2x$

$\rm \implies I_{min} = 5x - 4x$

$\rm \implies I_{min} =x$

Now , the ratio of the intensity of maxima to minima :-

$\rm \implies \dfrac{I_{max} }{I_{min} } = \dfrac{9x}{x}$

$\rm \implies \dfrac{I_{max} }{I_{min} } = \dfrac{9}{1}$

$\rm \implies \dfrac{I_{max} }{I_{min} } = 9 : 1$

Answer : 9:1