CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In a Young's double slit experiment the intensity of the resultant wave at a point P on the screen is I where the path difference between the waves from coherent sources $$S_1$$ and $$S_2$$ is $$\lambda$$. Then the intensity of the resultant wave at a point where the path difference is $$\lambda /4$$ is given by.


A
I/2
loader
B
2I
loader
C
4I
loader
D
I/2
loader

Solution

The correct option is D $$I/2$$
Given,
The intensity of the resultant wave at a point P on the screen is $$I$$ when the path difference is $$\lambda$$
Here we have to find the intensity of the resultant wave at a point where the path difference is $$\lambda/4$$.
We have Intensity, $$I = I $$ $$\cos^2\dfrac\phi2$$         ............... (1)
Where $$\phi$$ is phase difference
Also, Phase difference, $$\phi$$  $$ = \dfrac {2\pi}{\lambda}$$ X path difference      .................(2)
Given, path difference is $$\dfrac{\lambda}{4}$$
So, pahse difference, $$ \phi$$ = $$\dfrac{2\pi}{\lambda}$$ X $$\dfrac\lambda4$$
That is,                        $$\phi$$ = $$\dfrac\pi2$$ 
Substituting the value of $$\phi$$ in equation (1) we get,
  Intensity, $$I$$ = $$I$$ $$\cos^2\dfrac\pi4$$ 
                    


Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image