- #1

Andrew Tom

- 14

- 0

- 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 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.