Two coherent monochromatic light beams of intensities I and 4I are superimposed. The maximum and minimum possible intensities in the resulting beam are:
Answer & explanation
Correct answer: option 3
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 $\frac{A_1}{A_2}=\frac{\sqrt{I}}{\sqrt{4I}}=\frac{1}{2}$
$⇒\frac{I_{max}}{I_{min}}=\frac{9}{1}⇒I_{max}=9I,\,I_{min}=I$