In Young's double slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference $\lambda$ is $\mathrm{K}$ units. What is the intensity of light at a point where path difference is $\frac{\lambda}{3} ?$
Answer & explanation
Correct answer: option 2
$\text{ Let two intensities are equal to I , When path difference between waves is λ , the two waves are in phase}$
$ \text{Resultant Intensity is }I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}cos\phi$
$\text{ path difference is λ hence phase difference is }2\pi$
$\Rightarrow I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} = (\sqrt{I_1} + \sqrt{I_2})^2$
$\Rightarrow K = (\sqrt I + \sqrt I)^2 = 4I$
when path difference is $\frac{\lambda}{3}$ then phase difference is $\frac{2\pi}{3}$
$I'= I + I + 2 I cos \frac{2\pi}{3} = I = \frac{K}{4}$