Question

39.) In YDSE, d = 0.8 mm, a = 7200 Å and D = 2 m.
The minimum distance from central maximum at
which the average intensity is 50% of the
maximum is
(1) 0.45 mm
(2) 0.75 mm
(3) 1.25 mm
(4) 2.50 mm

Answer 1

Answer:

At phase Difference = π/2

path Difference = lemda /4

∆X = dy/D

dy/D = lemda /4

Y = 0.45mm

Answer 2

The minimum distance from central maximum is 0.45 mm.

(1) is correct option.

Explanation:

Given that,

Distance d= 0.8 mm

Wave length = 7200 A

Distance D = 2 m

The average intensity is 50% of the  maximum

We know that,

Resultant intensity at a point due to superposition of coherent waves of intensities $I_{1}$ and $I_{2}$ and phase difference $\phi$ can be expressed as

The resultant intensity is

$I_{R}=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos\phi$

Here, $I=I_{1}=I_{2}$

$I_{R}=4I\cos^2\dfrac{\phi}{2}$

At the position of maximum interference

$\phi=2n\pi$

The maximum intensity is

$I_{m}=4I$

So, the resultant intensity

$I_{R}=I_{max}\cos^2\dfrac{\phi}{2}$

Put the value into the formula

$\dfrac{I_{max}}{2}=I_{max}\cos^2\dfrac{\phi}{2}$

$\cos\dfrac{\phi}{2}=\dfrac{1}{\sqrt{2}}$

$\dfrac{\phi}{2}=\dfrac{\pi}{4}$

$\phi=\dfrac{\pi}{2}$

The phase difference at a point at which resultant intensity is 50% of maximum is $\dfrac{\pi}{2}$

We need to calculate the path difference

Using formula of path difference

$\Delta x=\dfrac{\lambda}{2\pi}\times\dfrac{\pi}{2}$

Put the value into the formula

$\Delta x=\dfrac{\labda}{4}$

We need to calculate the minimum distance from central maximum

Using formula of distance

$\dfrac{dy}{D}=\dfrac{\lambda}{4}$

$y=\dfrac{D\times\lambda}{4d}$

Put the value into the formula

$y=\dfrac{2\times7200\times10^{-10}}{4\times0.8\times10^{-3}}$

$y=0.45\ mm$

Hence, The minimum distance from central maximum is 0.45 mm.

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