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  • The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the distance between two fringe from the centre isOption:

The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the distance between two fringe from the centre is

Option: 1

2


Option: 2

1/2


Option: 3

4


Option: 4

16


Answers (1)

best_answer

Two waves of a single source having an amplitude A interfere. The resulting amplitude

\mathrm{A_{r}^{2}=A_{1}^{2}+A_{2}^{2}+2 A_{1} A \operatorname{Cos} \delta}
where  \mathrm{A_{1}=A_{2}=A\: and \: \delta=} phase difference between the waves
\mathrm{\Rightarrow \quad I_{r}=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \operatorname{Cos} \delta}

When the maxima occurs at the center,\mathrm{\delta=0}
\mathrm{\Rightarrow \quad \mathrm{I}_{\mathrm{r}_{1}}=4 \mathrm{I}}\quad \quad \ldots(1)
Since the phase difference between two successive fringes is \mathrm{2 \pi}, the phase difference between two points separated by a distance equal to one quarter of the distance between the two, successive fringes is equal to \mathrm{\delta=(2 \pi)\left(\frac{1}{4}\right)=\frac{\pi}{2}}  radius

\mathrm{\Rightarrow \quad \mathrm{I}_{\mathrm{r}_{2}}=4 \mathrm{l} \cos ^{2}\left(\frac{\pi / 2}{2}\right)=2 \mathrm{l}}\quad \quad \ldots(2)
using (1) and (2)

\mathrm{\frac{\mathrm{I}_{1}}{\mathrm{r}_{\mathrm{r}_{2}}}=\frac{4 \mathrm{I}}{2 \mathrm{I}}=2}
\therefore \quad \mathrm{A}
 

Posted by

SANGALDEEP SINGH

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