How far behind the pinhole should you place the viewing screen?

  • Thread starter Thread starter yo_man
  • Start date Start date
  • Tags Tags
    Screen
AI Thread Summary
To determine the distance behind the pinhole for the viewing screen, the half angular width of the central maximum is calculated using the formula lambda/d, where lambda is the wavelength (633 nm) and d is the pinhole diameter (0.12 mm). The relationship between the half angle, the radius of the central maximum (0.5 cm), and the distance to the screen is also utilized. The calculations involve converting units for consistency and solving for the distance x. The initial attempts at solving the equation led to incorrect results, indicating a need for careful unit conversion and application of the formulas. The final computed distance was approximately 0.95 meters, but further verification of the calculations is necessary.
yo_man
Messages
12
Reaction score
0

Homework Statement


You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm. You have a helium-neon laser (lambda=633 nm) and a 0.12-mm-diameter pinhole.
How far behind the pinhole should you place the viewing screen?

Homework Equations





The Attempt at a Solution



I'm not understanding how to go about doing this/starting this question
 
Physics news on Phys.org
Half angular width of the central maximum = lambda/d, where d is the slit width.
And half angle = radius of the central maximum/distance of the screen.
 
so I do 633 nm/1 cm --> of course I will convert to uniform units
and half angle is 0.5 cm/x

but how do I use these?
and should i be using the 1 cm or the .12 mm?
 
633*x^-9/0.12*10^-3 = 0.5*10^-2/x
 
Last edited:
so I did that and it's not right. so then I tried using 0.12 x 10^-3 instead of 0.06x 10^3
and I solved for x, and it's not right...
 
x = 0.5*10^-2*0.12*10^-3/633*10^-9 = 0.95m
 
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