Question

In Young's double slit experiment, using a monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$, is $K$ units. Then the intensity of light at a point where path difference is $\lambda /3$ ?

• A. $\frac{K}{2}$
• B. $2K$
• C. $4K$
• D. $\frac{K}{4}$

Answer 1

Answer: (D)

$\frac{K}{4}$

Resultant intensity at a point is given by $I_R=4I_o{\cos }^2\frac{\delta }{2}$
where $\delta$ is the phase difference and $I_o$ is the intensity of each light.
Case 1 : Path difference $\Delta x=\lambda$
Phase difference $\delta =\frac{2\pi }{\lambda }\times \Delta x=\frac{2\pi }{\lambda }\times \lambda =2\pi$
So, resultant intensity $I_1=4I_o{\cos }^2\pi =4I_o=K$
Case 2 : Path difference $\Delta x=\lambda /3$
Phase difference $\delta =\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}$
So, resultant intensity $I_2=4I_o\times co{s}^2\frac{\pi }{3}=K\times \frac{1}{2^2}=\frac{K}{4}$