Second bright fringe in Young's Experiment

[Andrew Tom]

Homework Statement

Second bright fringe in Young's Experiment

Relevant Equations

$n\lambda = \frac{xd}{D}$

In Young's double split experiment, a narrow beam of light of wavelength $589nm$ passes through two slits to form an interference pattern on a screen which is a perpendicular distance of $D$ metres away from the slits. The slit separation is $0.2mm$ and the second bright fringe is $6mm$ from the central maximum. Find $D$.

The formula given in the book is that the path difference is $\frac{xd}{D}$ where $x$ is the distance from the central maximum, $d$ is slit separation and $D$ is distance of screen from slits. So for bright fringes, $n\lambda = \frac{xd}{D}$ or $D=\frac{xd}{n\lambda}$. So for the second bright fringe, $n=1$ (since the first one is the central maximum at $n=0$). Hence $D=\frac{xd}{\lambda}$ which gives $D=2.04m$. However this is different from the answer at the back of the book.

 

[kuruman]

The proper way to count fringes is by considering the central maximum to be the zeroth maximum. When interference is constructive and you divide the path length difference from the slits by the wavelength, you get an integer. This integer is $n$ which is zero at the central maximum because the path lengths are equal. Thus the counting is
Central maximum $n = 0$
First two maxima $n = ± 1$
Second two maxima $n = ± 2$
$\dots$

I think you should use $n=2$.

 

[Andrew Tom]

The proper way to count fringes is by considering the central maximum to be the zeroth maximum. When interference is constructive and you divide the path length difference from the slits by the wavelength, you get an integer. This integer is $n$ which is zero at the central maximum because the path lengths are equal. Thus the counting is
Central maximum $n = 0$
First two maxima $n = ± 1$
Second two maxima $n = ± 2$
$\dots$

I think you should use $n=2$.

Thanks for your reply. Unfortunately this also gives a wrong answer (according to book) of D=1.01m.

There is a similar question which I am also getting the wrong answer for so I don't think it is a mistake in the book, however I can't see what I am doing wrong.

 

[kuruman]

Thanks for your reply. Unfortunately this also gives a wrong answer (according to book) of D=1.01m.

There is a similar question which I am also getting the wrong answer for so I don't think it is a mistake in the book.

D=1.01 m is the correct answer that you get when $n=2$. The formula is $D=\frac{xd}{n\lambda}$. With $n=1$, you got $D=2.04$ m; with $n=2$, you should get half as much because $n$ is in the denominator.