Question

Intensity of the central fringe in interference pattern is $0.01\;{\text{W}}/{{\text{m}}^2}$ then find intensity at a point having path difference $\lambda /3$ on screen from center in ${\text{mW}}/{{\text{m}}^2}$.
$\left( {\text{a}} \right){\text{ 2}}{\text{.5}}$
$\left( {\text{b}} \right){\text{ 5}}$
$\left( {\text{c}} \right){\text{ 7}}{\text{.5}}$
$\left( {\text{d}} \right){\text{ 10}}$

Answer 1

The above problem can be solved by using the principle of interference. There are two types of the fringes formed in the interference pattern, first one is the bright fringe and other one is the dark fringes. The intensity of the fringe varies with the distance from the screen, distance between the slits and phase angle.

Complete step by step answer:
Given:
The intensity of the central fringe is ${I_c} = 0.01\;{\text{W}}/{{\text{m}}^2}$.
The path difference at a point is $\Delta x = \lambda /3$.
The equation to calculate the intensity of reference fringe is given as:
$\Rightarrow {I_c} = 4{I_0}{\cos ^2}\left( {\phi /2} \right)$
Here, $\phi$ is the phase angle and its value for the central fringe is {\text{0^\circ }}.
Substitute $0.01\;{\text{W}}/{{\text{m}}^2}$for ${I_c}$ and {\text{0^\circ }}for $\phi$ in the equation which we have seen above to find the reference intensity.
$\Rightarrow 0.01\;{\text{W}}/{{\text{m}}^2} = 4{I_0}\left( {{{\cos }^2}\left( {0^\circ } \right)/2} \right)$
And on solving it, we get
$\Rightarrow {I_0} = 2.5 \times {10^{ - 3}}\;{\text{W}}/{{\text{m}}^2}$
The equation to calculate the phase angle for the point at which the intensity is to be find is,
$\Rightarrow \alpha = \dfrac{{2\pi }}{\lambda } \times \Delta x$
Substitute $\lambda /3$for $\Delta x$ in the above equation to find the phase angle at the point.
$\Rightarrow \alpha = \dfrac{{2\pi }}{\lambda }\left( {\dfrac{\lambda }{3}} \right)$
And on solving it, we get
$\Rightarrow \alpha = \dfrac{{2\pi }}{3}$
The equation to find the intensity at a point is given as:
$\Rightarrow I = {I_0}{\cos ^2}\left( {\dfrac{\alpha }{2}} \right)$
Substitute $\dfrac{{2\pi }}{3}$for $\alpha$ and $2.5 \times {10^{ - 3}}\;{\text{W}}/{{\text{m}}^2}$for ${I_0}$ in the above equation to find the intensity at the point.
$\Rightarrow I = 4\left( {2.5 \times {{10}^{ - 3}}\;{\text{W}}/{{\text{m}}^2}} \right)\left( {{{\cos }^2}\left( {\dfrac{{\dfrac{{2\pi }}{3}}}{2}} \right)} \right)$
And on solving it, we get
$\Rightarrow I = 2.5 \times {10^{ - 3}}\;{\text{W}}/{{\text{m}}^2}$
Now on changing the units, we get
$\Rightarrow I = \left( {2.5 \times {{10}^{ - 3}}\;{\text{W}}/{{\text{m}}^2}} \right)\left( {\dfrac{{1\;{\text{mW}}/{{\text{m}}^2}}}{{{{10}^{ - 3}}\;{\text{W}}/{{\text{m}}^2}}}} \right)$
And on solving it we get
$\Rightarrow I = 2.5\;{\text{mW}}/{{\text{m}}^2}$

Thus, the find intensity at a point having path difference $\lambda /3$ on screen from center is $2.5\;{\text{mW}}/{{\text{m}}^2}$and the option $\left( a \right)$ is the correct answer.

Note: Calculate the phase angle for the point at which the intensity has to be calculated. The calculated intensity at the point is in the ${\text{W}}/{{\text{m}}^2}$ but the options are given in the ${\text{mW}}/{{\text{m}}^2}$ so to find the correct option unit conversion is necessary.