Question

Intensity of central fringe in interference pattern is $0.1\dfrac{W}{{{m^2}}}$ then find intensity at a point having path difference $\dfrac{\lambda }{3}$ on screen from centre (in $\dfrac{W}{{{m^2}}}$)
A) $2.5$
B) $5$
C) $7.5$
D) $10$

Answer 1

Path difference is the measure of the path traveled by two waves. It is measured in terms of wavelength $\lambda$. If the measure of path difference between two waves is the even integer multiple of $\lambda$ means, it is called constructive interference. And in turn, if the measure of path difference is the odd multiple integrals of $\lambda$ means, it is called destructive interference.

Formula used:
$I = 4{I_o}{\cos ^2}\dfrac{\phi }{2}$
Where,
$I$- The intensity at any phase difference $\phi$
${I_o}$- Intensity of the light

Complete step by step answer:
(i) Intensity of the central fringe on the screen is given as $0.1\dfrac{W}{{{m^2}}}$. The intensity of the fringe can be found by the formula,
$I = 4{I_o}{\cos ^2}\dfrac{\phi }{2}$ --------- (1)
(ii) To find the value of intensity of point at particular path difference, we have to find the intensity of the light ${I_o}$ and the phase difference $\phi$
(iii) Therefore, in case of central fringe, $I = {I_C}$ and $\phi = 0$ applying these values in the equation (1)
$\Rightarrow {I_C} = 4{I_o}$ $\left[ {\cos 0 = 1} \right]$
$\Rightarrow {I_o} = \dfrac{{0.1}}{4}$ ---------- (2)
(iv)Finding the phase difference from the path difference by the relation $\phi = \dfrac{{2\pi }}{\lambda }\Delta x$ Where $\Delta x$ is the path difference it given here as $\dfrac{\lambda }{3}$. Therefore,
$\Rightarrow \phi = \dfrac{{2\pi }}{\lambda } \times \dfrac{\lambda }{3}$
$\Rightarrow \phi = \dfrac{{2\pi }}{3}$
(v) Now we found all the required values. Thus the value of the intensity of the point having path difference $\dfrac{\lambda }{3}$ is by the equation (1) is,
$I = 4{I_o}{\cos ^2}\dfrac{\phi }{2}$
$\Rightarrow I = 4 \times \dfrac{{0.1}}{4} \times {\cos ^2}\left( {\dfrac{{\tfrac{{2\pi }}{3}}}{2}} \right)$
$\Rightarrow I = 0.1 \times \dfrac{1}{4}$
$\Rightarrow I = 0.025W{m^{ - 2}}$

$\therefore I = 2.5mW{m^{ - 2}}$. Hence the correct option is A.

Note:
The phase difference between the waves can be found by finding the difference between the two-point at a wave. It is the measure of deviation between the two waves. The phase difference between the two waves will be the same if the two waves are moving together. This denotes that the crests of the two waves meet and the troughs too.