The interference pattern is obtained with two coherent light sources of intensity ratio $n$. In the interference pattern,the ratio $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ will be

  • A
    $\frac{\sqrt{n}}{(n + 1)^2}$
  • B
    $\frac{2\sqrt{n}}{(n + 1)^2}$
  • C
    $\frac{\sqrt{n}}{n + 1}$
  • D
    $\frac{2\sqrt{n}}{n + 1}$

Explore More

Similar Questions

An interference pattern is obtained with two coherent sources of intensity ratio $n:1$. The ratio $\frac{I_{\text{Max}}-I_{\text{Min}}}{I_{\text{Max}}+I_{\text{Min}}}$ will be maximum if

In an interference experiment,two coherent waves $S_1$ and $S_2$ are represented by $y_1 = 10 \sin(\omega t)$ and $y_2 = 10 \sin(\omega t - \pi/6)$ respectively. When these waves superimpose to form an interference pattern,the maximum intensity is ....... (Assume $K = 1$)

To produce the phenomenon of interference,we require two sources that emit radiation of:

Two monochromatic coherent light waves of amplitudes $A$ and $2A$,interfering at a point,have a phase difference of $60^{\circ}$. The intensity at that point will be proportional to (in $A^2$)

"If you illuminate two pinholes using two lamps, the interference pattern will not be observed" - Explain.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo