Question

Intensity of central fringe in interference pattern is $0.1W/m^2$ then find intensity at a point having path difference $\lambda /3$ on screen from centre ( in $mW/m^2$) :

• A. $2.5$
• B. $5$
• C. $7.5$
• D. $10$

Answer 1

The correct option is A $2.5$

Given that $I_c=0.1$$w/m^2$ be the intensify of centrical bright fringe.

Now intensity at any point with phase & is given by

$I=4I_o{\cos }^2\varphi /2$$\{I_o=$ intensity of light$\}$

Now, at central bright fringe $\varphi =0,I=I_c$

$=I_c=4I_o{\cos }^{2}0$

$=\frac{0.1}{4}=I_o$

Now at path difference $\Delta x=\frac{\lambda }{3}$

as we know $\varphi =k\Delta x$

& $k=\frac{2\pi }{\lambda }=$ angular wave number

$\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}$

$I=4I_o{\cos }^2\left(\frac{2\pi /3}{2}\right)$

$\Rightarrow I=4\times \frac{0.1}{4}\times \frac{1}{4}=0.025w/m^2$

$I=2.5mW/m^2$.