Question

In a double-slit experiment, the separation between the slits is $d=0.25\, cm$ and the distance of the screen $D=100 \,cm$ 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 $=\frac{x d }{ D } \Rightarrow$ 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 $=\frac{\pi}{3} \Rightarrow \phi=60^{\circ}$
$I _{ R }= I _1+ I _2+2 \sqrt{ I _1 I _2} \cos 60^{\circ} \Rightarrow I _{ R }= I + I +2 I \times \frac{1}{2}$
$I _{ R }=3 I \Rightarrow I _{ R }=\frac{3 I _0}{4}$