Find ratio of brightness & the distance the 6th dark fringe is from central fringe

AI Thread Summary
The discussion focuses on calculating the brightness ratio and the distance of the sixth dark fringe from the central fringe using relevant equations. The initial calculations for part (a) were found to be incorrect, prompting a review of the formulas and units used, particularly regarding the angle β being in radians. The user identified an error in their calculations and corrected part (a), while still struggling with part (b). It was emphasized that attention to units and calculator settings is crucial for accuracy. Ultimately, the user confirmed they achieved the correct answer after addressing these issues.
hraghav
Messages
48
Reaction score
5
Homework Statement
Light of wavelength λ = 517.4× 10^−9 m is incident on a metal plate that has two thin, parallel slits etched into it a distance of d = 5.44 × 10^−6 m apart. A distance of L=1.29 m away lies a viewing panel onto which the diffraction pattern can be displayed.
The metal plate with the two slits is replaced with a metal plate with a single slit that has a width of a= 3.96 × 10^−6 m. a) What is the ratio of the brightness at y = 1.26 m (from the central fringe) to the brightness of the central fringe?

b) The metal plate with the two slits is replaced with a metal plate with a single slit that has a width of a = 7.96 × 10^-6 m. How far from the central fringe is the 6th dark fringe (destructive interference)?
Relevant Equations
a) I = Io(sinβ / β)^2
β = (pi * a*sinθ) / λ
sinθ = y / L

b) asinθ=mλ
a) sinθ = y / L where y = 1.26 m and L = 1.29m
sinθ = 0.9767

β = (pi * a*sinθ) / λ where λ = 517.4 * 10^-9 m and a= 3.96 × 10^−6 m
β = 0.000012151 / 517.4 * 10^-9 = 23.485

I / Io = (sin(23.485) / 23.485)^2
I / Io = 0.0002879
which is 0.2879 * 10^-3

b) asinθ=mλ where m = 6 and λ = 517.4 * 10^-9 m
sinθ = 0.000003104 / a where a = 7.96 × 10^-6 m
sinθ = 0.39

y = Lsinθ. where L = 1.29 m
y = 0.5031 m
which is 50.31 * 10^-2 m

Both these answers are incorrect. Could someone please let me know where am I making an error?
 
Physics news on Phys.org
hraghav said:
Relevant Equations:
sinθ = y / L
Do you have the correct trig function here?
 
hraghav said:
Homework Statement:
Relevant Equations: a) I = Io(sinβ / β)^2
β = (pi * a*sinθ) / λ
When you use this formula, is ##\beta## in degrees or in radians?
 
TSny said:
When you use this formula, is ##\beta## in degrees or in radians?
I got my error in part a but still can't figure out part b
 
When you use the formula for obtaining the angle ##\beta##, the angle will be in units of radians. So, be careful with the units.

Did you catch your mistake in writing ##\sin \theta = y/L##?

EDIT: Never mind, you said you got part (a) correct. So, you must have made these corrections. In part (b) did you make the correction in ##y = L \sin \theta##? Also, be careful if you've been switching your calculator back and forth between radian and degree mode.
 
TSny said:
When you use the formula for obtaining the angle ##\beta##, the angle will be in units of radians. So, be careful with the units.

Did you catch your mistake in writing ##\sin \theta = y/L##?

EDIT: Never mind, you said you got part (a) correct. So, you must have made these corrections. In part (b) did you make the correction in ##y = L \sin \theta##? Also, be careful if you've been switching your calculator back and forth between radian and degree mode.
Yes I understood what you mean, Thanks I got the correct answer
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top