# Second bright fringe in Young's Experiment

• Andrew Tom
In summary, the Young's double split experiment involves a narrow beam of light passing through two slits to form an interference pattern on a screen. The distance ##D## between the slits and the screen can be calculated using the formula ##D=\frac{xd}{n\lambda}##, where ##x## is the distance from the central maximum, ##d## is the slit separation, and ##n## is the number of the bright fringe. The proper way to count fringes is by considering the central maximum to be the zeroth maximum, and the second bright fringe corresponds to ##n=1##. Therefore, for the second bright fringe, the correct value of ##n## is ##n=2##, which gives

#### 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.

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##.

kuruman said:
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##.