Second bright fringe in Young's Experiment

  • Thread starter Thread starter Andrew Tom
  • Start date Start date
  • Tags Tags
    Experiment Fringe
AI Thread Summary
In Young's double-slit experiment, a light beam of wavelength 589 nm creates an interference pattern with a slit separation of 0.2 mm. The second bright fringe appears 6 mm from the central maximum, leading to calculations for the distance D from the slits to the screen. The correct formula for D is D = (xd)/(nλ), where n is the fringe number. For the second bright fringe, n should be 2, resulting in D = 1.01 m, which aligns with the book's answer. Misunderstandings arose from incorrectly counting the central maximum as n=1 instead of n=0.
Andrew Tom
Messages
14
Reaction score
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.
 
Physics news on Phys.org
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##.
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.
 
Andrew Tom said:
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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top