The intensity at the maximum in a Young's double slit experiment … - Vidyadip

MCQ

The intensity at the maximum in a Young's double slit experiment is $I_0$. Distance between two slits is $d = 5\lambda,$ where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10\,d\,?$

A
$\frac{{{I_0}}}{4}$
B
$\;\frac{3}{4}{I_0}$
✓
$\;\frac{{{I_0}}}{2}$
D
$\;{I_0}$

✓

Answer

Correct option: C.

$\;\frac{{{I_0}}}{2}$

c
Here, $d=5 \lambda$, $D=10\,d$, $y=\frac{d}{2}$ .

Resultant Intensity at $y=\frac{d}{2}, I_{y}=?$

The path difference between two waves at $y=\frac{d}{2}$

$\Delta x = d\tan \theta = $ $d \times \frac{y}{D} = $ $\frac{{d \times \frac{d}{2}}}{{10d}} = $ $\frac{d}{{20}} = \frac{{5\lambda }}{{20}} = \frac{\lambda }{4}$

Corresponding phase difference, $\phi=\frac{2 \pi}{\lambda} \Delta x=\frac{\pi}{2}$

Now, maximum intensity in Young's double slit experiment,

${I_{\max }} = {I_1} + {I_2} + 2{I_1}{I_2}$

${I_0} = 4I$ $(\because \,\,{I_1}\, = \,{I_2}\, = \,I)$

$\therefore I=\frac{I_{0}}{4}$

Required intensity,

${I_y} = {I_1} + {I_2} + 2{I_1}{I_2}\cos \frac{\pi }{2}$ $ = 2I = \frac{{{I_0}}}{2}$

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