Question

A point is situated at 6.5 cm and 6.65 cm from two coherent sources. Find the nature of illumination at the point if wavelength of light is $5000{}^ \circ A$

Answer 1

Hint: Calculate the path difference $\Delta x = n \times \lambda $. If the value of the path difference is an even multiple of $\dfrac{\lambda }{2}$ then it is a constructive interference and the point will be bright. If the value of the path is an odd multiple of $\dfrac{\lambda }{2}$ then it is a destructive interference and the point will be dark.

Complete step by step answer:
We are given a point is situated at 6.5 cm and 6.65 cm from two coherent sources and wavelength of the light is $5000{}^ \circ A$
We have to find the nature of illumination at this point.
The nature of illumination can be said in terms of constructive interference or destructive interference.
If the point has constructive interference then the point will be bright, if the point has destructive interference then the point will be dark.
So, to find out the nature, first calculate the path difference between the given two sources.
If the value of the path difference is an even multiple of $\dfrac{\lambda }{2}$ then it is a constructive interference and the point will be bright. If the value of the path is an odd multiple of $\dfrac{\lambda }{2}$ then it is a destructive interference and the point will be dark.
$
  \Delta x = n \times \lambda \\
\implies \Delta x = {x_2} - {x_1} \\
  {x_2} = 6.65cm,{x_1} = 6.5cm \\
 \implies \Delta x = 6.65 - 6.5 = 0.15cm \\
\implies \Delta x = 15 \times {10^{ - 4}}m \\
 \implies \lambda = 5000{}^ \circ A = 5 \times {10^{ - 7}}m \\
\implies \Delta x = n \times \lambda \\
 \implies 15 \times {10^{ - 4}} = n \times 5 \times {10^{ - 7}} \\
\implies n = \dfrac{{15 \times {{10}^{ - 4}}}}{{5 \times {{10}^{ - 7}}}} \\
  n = 3 \times {10^3} = 3000 \\
 $
Substitute the value of n in $\Delta x = n \times \lambda $ to find the path difference in terms of lambda.
$
  \Delta x = n \times \lambda \\
\implies \Delta x = 3000 \times \lambda \\
\implies \Delta x = 6000 \times \dfrac{\lambda }{2} \\
\therefore \Delta x = 6000\dfrac{\lambda }{2} \\
 $
The value of the path difference is an even multiple of $\dfrac{\lambda }{2}$
So, the nature of the illumination is constructive interference and the point will be bright.

Note:
Here the term ‘illumination’ is given. Illuminance is measured as the amount of light which is striking the surface of an object or a body whereas luminance is the amount of light reflected from the surface which is illuminated. Illuminance is the incident light and luminance is the reflected light. There is a slight difference between luminance and illuminance. So do not confuse between these two terms.