Question

In a Young's double slit experiment
$\lambda =500nm,d=1mm\mathrm{and}D=1m.$
The minimum distance from the central maximum for which the intensity is half of the maximum intensity is

• A. $2\times {10}^{-4}m$
• B. $1.25\times {10}^{-4}m$
• C. $4\times {10}^{-4}m$
• D. $2.5\times {10}^{-4}m$

Answer 1

The correct option is B $1.25\times {10}^{-4}m$

Intensity of central maxima $I^{}={\left(2A_0\right)}^2=4{A_0}^2=4I_0$
Intensity at distance $y$ from the central maxima is half of the maximum intensity if $I_0=4I_{0}co{s}^2\frac{\varphi }{2}=\frac{4I_0}{2}$

If
${\cos }^2\frac{\varphi }{2}=\frac{1}{2}\Rightarrow \frac{\varphi }{2}=\frac{\pi }{4}\Rightarrow \varphi =\frac{\pi }{2}$
Now we know
$\varphi =\frac{2\pi }{\lambda }\Delta x$
where $\Delta x$ is the path difference between the two interfering waves
also $\Delta x=\frac{yd}{D}$
$\Rightarrow \varphi =\frac{2\pi }{\lambda }\frac{yd}{D}=\frac{\pi }{2}$
$\Rightarrow y=\frac{\lambda D}{4d}\Rightarrow \frac{500\times {10}^{-9}\times 1}{4\times {10}^{-3}}=1.25\times {10}^{-4}m$