Question

In a double-slit experiment, the separation between the slits is $d=0.25cm$ and the distance of the screen $D=100cm$ from the slits. If the wavelength of light used is $\lambda =6000\mathring{A}$ and $I_0$ is the intensity of the central bright fringe, the intensity at a distance $x=4\times 10^{-5}m$ from central maximum is

• A. $I_0$
• B. $\frac{I_0}{2}$
• C. $\frac{3I_0}{4}$
• D. $\frac{I_0}{3}$

Answer 1

Answer: (C)

Path difference $=\frac{xd}{D}\Rightarrow$ Path difference $=\frac{4\times 10^{-5}\times 0.25\times 10^{-2}}{1}$
Path difference $=1\times 10^{-7}$
Phase difference $=\frac{\text{ path diff. }}{\lambda }\times 2\pi \Rightarrow$ Phase difference $=\frac{1\times 10^{-7}}{6\times 10^{-7}}\times 2\pi$
Phase difference $=\frac{2\pi }{6}\Rightarrow$ Phase difference $=\frac{\pi }{3}\Rightarrow \varphi =60^{\circ }$
$I_R=I_1+I_2+2\sqrt{I_1{I}_2}\cos 60^{\circ }\Rightarrow I_R=I+I+2I\times \frac{1}{2}$
$I_R=3I\Rightarrow I_R=\frac{3I_0}{4}$