Question

Consider a Michelson interferometer as shown in the figure below. When the wavelength of the laser light source is switched from 400 nanometer to 500 nanometer, it is observed that the intensity measured at the output port P goes from a minimum to a maximum. This observation is possible when the smallest path difference between the two arms of the interferometer is nanometer.

• A. 1000

Answer 1

The correct option is A 1000
As per the question the optical path difference in both the cases $\left(400\text{nm}\text{or}500\text{nm}\right)$ should be same but for $400\text{nm}$ wavelength we should get distructive interference whereas for $500\text{nm}$ wavelength we should get constructive interference.
$\therefore$ For constructive interference optical path difference $\left(2x\right)=n{\lambda }_c$
For distructive interference optical path difference
$\left(2x\right)=\frac{(2m+1){\lambda }_0}{2}$
As per question,
$h{\lambda }_c=\frac{(2m+1){\lambda }_0}{2}$
$n\left(500\right)=\frac{(2m+1)400}{2}$
$\therefore$ For $n=2$ or $m=2$ condition satisfied
$\therefore$ Optical path difference
$2x=n\lambda =2\left(500\text{nm}\right)=1000\text{nm}$