Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Wave Optics

Question:

Two coherent monochromatic light beams of intensities I and 4I are superimposed. The maximum and minimum possible intensities in the resulting beam are:

Options:

5 I and I

5 I and 3I

9 I and I

9 I and 3 I

Correct Answer:

9 I and I

Explanation:

Intensity ∝ (Amplitude)2

$⇒ I ∝ A^2$

When two waves (beams) of amplitude $A_1$ and $A_2$ superimpose, at maxima and minima, the amplitude of the resulting wave are $(A_1 + A_2)$ and $(A_1 - A_2)$ respectively. If the maximum and minimum possible intensities are $I_{max}$ and $I_{min}$ respectively, then

$I_{max}∝(A_1+A_2)^2$

And 

$I_{min}∝(A_1-A_2)^2$

$⇒\frac{I_{max}}{I_{min}}=(\frac{A_1+A_2}{A_1-A_2})^2=\{\frac{\frac{A_1}{A_2}+1}{\frac{A_1}{A_2}-1}\}$ where $\frac{A_1}{A_2}=\frac{\sqrt{I}}{\sqrt{4I}}=\frac{1}{2}$

$⇒\frac{I_{max}}{I_{min}}=\frac{9}{1}⇒I_{max}=9I,\,I_{min}=I$