Physics, asked by vaishnavi3528, 9 months ago

A monochromatic source of wavelength 600nm was used in YSD exp.to produce interference pattern.I is the intensity of light at a point on the screen where the path diff is 150nm.The intensity of light at a point where the path diff is 200nm is given by?​

Answers

Answered by CarliReifsteck
0

Given that,

Wavelength = 600 nm

Intensity is I, where the path difference is 150 nm

We know that,

The formula of intensity is

I=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos\theta

We need to calculate the value of θ

Using formula of path difference

\theta=2\pi\times\dfrac{\phi}{\lambda}

Where, \phi = path difference

\lambda = wave length

Put the value into the formula

\theta=2\pi\times\dfrac{150}{600}

\theta=1.57

We need to calculate the intensity

Using equation (I)

I=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos\theta

Here, I_{1}=I_{2}=I_{0}

Put the value of θ

I=2I_{0}+2\sqrt{I_{0}^2}\cos(1.57)

I=4I_{0}

We need to calculate the value of of θ

Using formula of path difference

\theta=2\pi\times\dfrac{\phi}{\lambda}

Put the value into the formula

\theta=2\pi\times\dfrac{200}{600}

\theta=2.09

We need to calculate the intensity

Using equation (I)

I'=I_{0}+I_{0}+2\sqrt{I_{0}I_{0}}\cos\theta

Put the value of θ

I'=2I_{0}+2\sqrt{I_{0}^2}\cos(2.09)

I'=4I_{0}

Put the value of I_{0}

I'=4\times\dfrac{I}{4}

I'=I

Hence, The intensity of light at a point will be I.

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