Calculating Wave Length in Monochromatic Light

AI Thread Summary
The discussion centers on calculating the wavelength of monochromatic light based on given parameters, including the distance to the screen and the position of the dark fringe. The initial calculation yielded a wavelength of approximately 429.427 nm, but it was noted that the answer should be rounded to 430 nm for consistency with the significant figures of the measurements. A critical correction was made regarding the identification of the dark fringe; the fourth dark fringe corresponds to m=3.5, not m=4, which affects the wavelength calculation. The use of the small angle approximation was also highlighted, where sin(theta) can be approximated as tan(theta) when the angle is small. This discussion emphasizes the importance of accurately identifying fringe positions in wave interference problems.
Kris1120
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Homework Statement


A screen is illuminated by monochromatic
light.
The distance from the slits to the screen is
7.4 m.
What is the wave length if the distance from the central bright region to the fourth
dark fringe is 2.6 cm.
Answer in units of nm.


Homework Equations


x=L*tan(theta)

sin(theta)=(m+.5) lambda/d


The Attempt at a Solution



theta= inverse tan( 2.6e-2 / 7.4) = .201309

lambda = (.55e-3) sin(.201309) / 4.5 = 429.427 nm
 
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correct, but way too many digits. 430 nm is the proper way to give the answer, since only two digits are given for L and x.

By the way, when the angle is very small (such as is the case here) sin theta = tan theta.

That way, you can save some time and see that x/L=(m lambda)/d

It'll be like that whenever L is much greater than x
 
Kris1120 said:

The Attempt at a Solution



theta= inverse tan( 2.6e-2 / 7.4) = .201309

lambda = (.55e-3) sin(.201309) / 4.5 = 429.427 nm


I don't believe this is correct. The fourth dark fringe does not correspond to m=4 in that equation.
 
Alphysicist is right. Fourth dark fringe is halfway between the 3rd & 4th bright fringe, so the "m-factor" should be 3.5.
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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