question_answer

The intensity at the maximum in a Young's double slit experiment is \[{{\text{I}}_{\text{0}}}\text{.}\]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 d?                                                                  

A)  \[{{I}_{0}}\]                                      

B)  \[\frac{{{I}_{0}}}{4}\]

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

D)   v

Correct Answer: D

Solution :

                 Path difference \[={{S}_{2}}P-{{S}_{1}}P\] \[=\sqrt{{{D}^{2}}+{{d}^{2}}}-D\] \[=D\left( 1+\frac{1}{2}\frac{{{d}^{2}}}{{{D}^{2}}} \right)-D\] \[=D\left[ 1+\frac{{{d}^{2}}}{2{{D}^{2}}}-1 \right]=\frac{{{d}^{2}}}{2D}\] \[\Delta x=\frac{{{d}^{2}}}{2\times 10d}=\frac{d}{20}=\frac{5\lambda }{20}=\frac{\lambda }{4}\] \[\Delta \,\text{o}|\,=\frac{2\pi }{\lambda }.\frac{\lambda }{4}=\frac{\lambda }{2}\] So, intensity at the desired point is \[I={{I}_{0}}{{\cos }^{2}}\frac{\text{o }\!\!|\!\!\text{ }}{2}={{I}_{0}}{{\cos }^{2}}\frac{\pi }{4}=\frac{{{I}_{0}}}{2}\]