Single slit diffraction - distance between 1st&2nd order dark fringes

AI Thread Summary
Monochromatic light with a wavelength of 580 nm passes through a single slit measuring 2.60 µm wide, with a screen positioned 2.10 m away. The distance between the first- and second-order dark fringes is calculated using the formula y=(n*lambda*L)/d, resulting in y1=0.468 m and y2=0.937 m, leading to a difference of 0.468 m. Clarification is sought on the meaning of variables in the formula and the conditions for its approximation. The discussion emphasizes the importance of calculating the angle of deviation for accurate results, particularly when the angle is small enough to use the approximation of sin(theta) with tan(theta). Understanding these concepts is crucial for solving single slit diffraction problems effectively.
louza8
Messages
58
Reaction score
0

Homework Statement


Monochromatic light with wavelength 580nm passes through a single slit 2.60um wide and 2.10m from a screen.

Find the distance between the first- and second-order dark fringes on the screen.

Homework Equations



y=(n*lamda*L)/d

The Attempt at a Solution



y1=580nm*2.1m/2.1um=0.468m
y2=580nm*2.1m*2/2.1um=0.937m

y2-y1=0.468m

I am not sure how to handle the dark fringes in single slit diffraction, so I am not sure where I am going wrong.

Thanks for people's help in advance.
 
Physics news on Phys.org
Do you know what the letters mean in the formula y=(n*lambda*L)/d and when is it a good approximation?

ehild
 
y is the distance between the fringes hitting the screen, n is the order, lambda is the wavelength, L is the distance to the screen, d is the width of the slit?

i don't know when it is a good approximation or what you mean by that.
 
Yo know a formula which contain the angle of deviation from the original direction of the light ray, hitting the slit. It is the exact one. ehild
 
n*lamda = d*sin(theta) ?

so would i use the angle found theta, and the known length L in a triangle to find the difference in y?
 
Yes. Calculate theta for n=1 and for n=2. The relation between L and y is y/L=tan(theta). If theta was very small, not more than a few degrees, you could approximate sin(theta) with tan(theta) and the equation in your first post would be valid. It is not the case now.

ehild
 
thanks ehild, helpful as always
 
Back
Top