question_answer

The intensity at the maximum in a Young's double slit experiment is \[{{I}_{0}}\]. Distance between two slits is \[d=5\lambda ,\] where \[\lambda \] the wavelength of light is used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance \[D=10\text{ }d\]?                                                                          [NEET - 2016]

A)  \[{{I}_{0}}\]               

B)  \[\frac{{{I}_{0}}}{4}\]

C)  \[\frac{3}{4}{{I}_{0}}\]          

D)                   v

Correct Answer: D

Solution :

[d] Path difference

\[={{S}_{2}}P-{{S}_{1}}P\]

\[=\sqrt{{{D}^{2}}+{{d}^{2}}}-D\]

\[=D\left( 1+\frac{1}{2}\frac{{{d}^{2}}}{{{D}^{2}}} \right)-D\]

\[=D\left[ 1+\frac{{{d}^{2}}}{2{{D}^{2}}}-1 \right]=\frac{{{d}^{2}}}{2D}\]

\[\Delta x=\frac{{{d}^{2}}}{2\times 10d}=\frac{d}{20}=\frac{5\lambda }{20}=\frac{\lambda }{4}\]

\[\Delta \,o|\,=\frac{2\pi }{\lambda }.\frac{\lambda }{4}=\frac{\lambda }{2}\]

So, intensity at the desired point is

\[I={{I}_{0}}{{\cos }^{2}}\frac{\phi }{2}={{I}_{0}}{{\cos }^{2}}\frac{\pi }{4}=\frac{{{I}_{0}}}{2}\]