Questions & Answers

Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible resulting intensities are :
A. 5I and 0
B. 5I and 3I
C. 9I and I
D. 9I and 3I

Answer Verified Verified
Hint: Take ${{I}_{1}}$ and ${{I}_{2}}$ as the two intensities of two coherent monochromatic light beams, then find out the maximum and minimum intensity by constructing a formula with the help of those two assumed variable.

Complete step by step answer:
If ${{I}_{1}}$ and ${{I}_{2}}$ are two intensities of two coherent monochromatic light beams, then
${{I}_{\max }}$=($\sqrt{{I_1}}$ + $\sqrt{{I_2}}$)$^{2}$
${{I}_{\min }}$=($\sqrt{{I_1}}$ - $\sqrt{{I_2}}$)$^{2}$
Now substituting the value of ${{I}_{1}}$ and ${{I}_{2}}$ with I and 4I,
Therefore, maximum intensity
${{I}_{\max }}$=($\sqrt{I}$+$\sqrt{4I}$)$^{2}$
On solving it comes,
${{I}_{\max }}$=9I
Therefore, minimum intensity,
${{I}_{\min }}$=($\sqrt{I}$-$\sqrt{4I}$) $^{2}$
On solving it comes,
${{I}_{\min }}$=I
Option C is the correct option.

Additional information:
Laser light is an example of a source of coherent monochromatic light.Two waves are coherent if the phase difference between them is constant. For this to be the case they must have the same frequency. Monochromatic means having only one wavelength of light present, it is actually a source of light which has constant phase difference and same frequency
Intensity is the power transferred per unit area, where area is measured on the plane perpendicular to the direction of propagation of the energy. In the SI system, it has units watts per square metre.

Maximum intensity is =($\sqrt{I}$+$\sqrt{4I}$)$^{2}$
Minimum intensity is=($\sqrt{I}$-$\sqrt{4I}$)$^{2}$
The light beams are coherent.
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