Single Slit Diffraction and Monochromatic light

AI Thread Summary
Monochromatic light with a wavelength of 419 nm passes through a single slit, creating a diffraction pattern on a screen 88 cm away. The distance to the first-order dark band is 0.29 cm from the center, leading to the need to calculate the slit width. The correct approach involves using the formula for the first minimum, sin θ = nλ/a, where 'a' is the slit width. Calculating the angle θ using arctan(0.29/88) is essential for accurate results. The initial calculations led to discrepancies, but following the suggested formula and method yields the correct slit width.
jones268
Messages
4
Reaction score
0
Monochromatic light with a wavelength of 419 nm passes through a single slit and falls on a screen 88 cm away. If the distance of the first-order dark band is 0.29 cm from the center of the pattern, what is the width of the slit?

My knowns are as follows:
L= 88 cm (distance from slit to screen)
λ= 419 nm
WCM= 0.58 cm (width of the central max, I assumed it was twice the distance from the center of the patter to the first order dark band)
w=? unknown width of the slit

I thought I should use the following equation:

WCM=[(2)(λ)(L)]/[(square root of : (w^2 - λ^2))]

I plugged the numbers into the equation and solved for w, but came up with the wrong answer, I'm not quite sure what I'm doing wrong...
 
Physics news on Phys.org
you should be able to solve for an angle using the distance of the first order band. After that, there will be a formula that relates all of your other components to theta and slit width.
 
Shouldn't you be using the formula for the direction to the first minimum
sin θ = nλ/a where a is the slit width?
 
Thanks stonebridge for the help, I never thought of using that equation for L. But I'm still coming up short on the right answer. Instead of 0.0127145 cm as my answer, I'm coming up with 0.0165528450 cm. :/ This is so frustrating.
 
use the formula that stonebridge just gave you. I got the right answer. make sure that you calculate your theta using arctan(.29/88)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top