Question

(a) (i)'Two independent monochromatic sources of light cannot produce a sustained interference pattern'. Give reason.
(ii) Light waves each of amplitude "a" and frequency " $\omega^{\prime \prime},$ emanating from two coherent light sources superpose at a point. If the displacements due to these waves is given by $y_{1}=a \cos \omega t$ and $y_{2}=\operatorname{acos}(\omega t+\phi)$ where $\phi$ is the phase difference between the two, obtain the expression for the resultant intensity at the point.
(b) In Young's double slit experiment, using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda,$ is $\mathrm{K}$ units. Find out the intensity of light at a point where path difference is $\lambda / 3$.

Answer 1

Hint: When monochromatic light going through two limited slits enlightens a distant screen, a trademark example of light and dark fringes is observed. This obstruction design is brought about by the superposition of covering light waves beginning from the two slits. This is known as the Young's double slit experiment which is needed for solving the question.

Complete step by step answer: From the given question, we can determine that
 (a)(i) Two independent monochromatic sources of light cannot produce a sustained interference because:
(1) If the sources are not coherent, they cannot emit waves continuously.
(2) Independent sources emit the waves, which don't have the same phase or a constant phase difference.
(ii) given $\mathrm{y}_{1}=\mathrm{a} \cos \omega \mathrm{t}$
$\mathrm{y}_{2}=\mathrm{a} \cos (\omega \mathrm{t}+\phi)$by superposition principle, resultant displacement, $\mathrm{y}=\mathrm{y}_{1}+\mathrm{y}_{2}$
$\Rightarrow$ $\mathrm{y}=\mathrm{a} \cos \omega \mathrm{t}+\mathrm{a} \cos (\omega \mathrm{t}+\phi)$
$\Rightarrow$ $\mathrm{y}=2 \mathrm{a} \cos (\phi / 2) \cdot \cos (\omega \mathrm{t}+\phi / 2)$
$\Rightarrow$ $\mathrm{y}=\mathrm{A} \cos (\omega \mathrm{t}+\phi / 2)$
it is an equation of simple harmonic plane progressive wave, whose amplitude is $\mathrm{A}$, here $\mathrm{A}=2 \mathrm{a} \cos (\phi / 2)$
Now intensity is proportional to square of amplitude, therefore
$\mathrm{I}=\mathrm{KA}^{2}=4 \mathrm{Ka}^{2} \cos ^{2}(\phi / 2)$ where $\mathrm{K}$ is proportionality constant.
(b) In interference the intensity I at a point is given by,
$\mathrm{I}=\mathrm{I}_{0} \cos ^{2}(\pi / \lambda) \mathrm{x}$ where $\mathrm{x}=$ path difference,
$\lambda =\text{ }$ wavelength
${{\text{I}}_{0}}=\text{ }$ intensity of central maximum
when $x=\lambda, I=K$
$\mathrm{K}=\mathrm{I}_{0} \cos ^{2}(\pi / \lambda) \lambda$ or $\quad \mathrm{K}=\mathrm{I}_{0} \cos ^{2} \pi=\mathrm{I}_{0}$
when $x=\lambda / 3, I=I^{\prime}$
$\mathrm{I}=\mathrm{I}_{0} \cos ^{2}(\pi / \lambda) \lambda / 3$ or $\mathrm{I}^{\prime}=\mathrm{I}_{0} \cos ^{2}(\pi / 3)=\mathrm{I}_{0}(1 / 2)^{2}=\mathrm{K} / 4$

Note: We must thoroughly understand the various concepts of the Young’s double slit experiment. Also, we must keep the concept of coherence in mind.