Question #119703
In young’s double slit experiment the intensity of light at a point on the screen where the path difference is is K. The intensity of light at a point where the path difference is [ is the wave length of light used] is
1
Expert's answer
2020-06-04T09:55:08-0400

The condition of the problem is:

In young’s double slit experiment the intensity of light at a point on the screen where the path difference is λ is K. The intensity of light at a point where the path difference is λ/4 is...

Solution

For path difference of one wavelength, the phase difference is


ϕ1=2πλx=2πλλ=2π.\phi_1=\frac{2\pi}{\lambda}x=\frac{2\pi}{\lambda}\lambda=2\pi.


For x=λ/4 the phase difference is


ϕ2=2πλx=2πλλ4=π2.\phi_2=\frac{2\pi}{\lambda}x=\frac{2\pi}{\lambda}\frac{\lambda}{4}=\frac{\pi}{2}.

The intensity of light is


I=K cos2(ϕ2).I=K\text{ cos}^2\bigg(\frac{\phi}{2}\bigg).

Therefore,


I1=K,I2=K cos2(π/22)=K2.I_1=K, \\I_2=K\text{ cos}^2\bigg(\frac{\pi/2}{2}\bigg)=\frac{K}{2}.

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