Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Wave Optics

Question:

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 $\frac{\lambda}{3} ?$

Options:

\(\frac{K}{6}\)

\(\frac{K}{4}\)

\(\frac{K}{2}\)

\(\frac{K}{8}\)

Correct Answer:

\(\frac{K}{4}\)

Explanation:

$\text{ Let two intensities are equal to I , When path difference between waves is λ , the two waves are in phase}$

$ \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 $\frac{2\pi}{3}$

$I'= I + I + 2 I cos \frac{2\pi}{3} = I = \frac{K}{4}$