Question

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?​

Answer 1

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.