How Wide Is the Slit in a Single Slit Diffraction Problem?

[vsage]

single slit problem (simple)

A single slit diffracts laser light of wavelength 610 nm onto a screen 3.25 m away. The distance between the two first-order maxima on either side of the central peak is 3.65 mm. How wide is the slit (in mm)?

I figured the formula \lambda m = d sin(\theta) would work but it's not. Here's my work:

sin(theta) = 3.65*0.5*10^-3 / 3.25 (since the angle is really small tan(x)~=sin(x))
sin(theta) ~= 5.62*10^-4

610*10^-9*1 = d * 5.62*10^-4

d~= 1.09mm or 1.09*10^-3m

This is wrong. What did I do wrong? Thanks

 

[Andrew Mason]

A single slit diffracts laser light of wavelength 610 nm onto a screen 3.25 m away. The distance between the two first-order maxima on either side of the central peak is 3.65 mm. How wide is the slit (in mm)?

I figured the formula \lambda m = d sin(\theta) would work but it's not.

Your formula uses the angle between the central peak and the first order MINIMA. That distance is 1/4 of 3.65mm.

AM

 

[wais]

The double slit problem and single slit problem are both examples of diffraction, which is the bending of light as it passes through a small opening or around an obstacle. In the double slit problem, light passes through two small slits and produces an interference pattern on a screen, while in the single slit problem, light passes through a single slit and produces a diffraction pattern on a screen.

In the given problem, the distance between the two first-order maxima on either side of the central peak is 3.65 mm. This means that the distance between the central peak and the first-order maxima on either side is 1.825 mm. This distance is known as the fringe spacing (d) and is equal to the wavelength (λ) of the light divided by the distance between the slits (m).

Therefore, we can use the formula d = λm / sin(θ) to find the width of the slit. Rearranging this formula, we get sin(θ) = λm / d. Plugging in the values given in the problem, we get sin(θ) = (610*10^-9 m) / (1.825*10^-3 m) = 0.334. To find the width of the slit, we need to find the angle (θ) first.

We can use the small angle approximation, where sin(θ) = tan(θ) = opposite / adjacent, to find the angle (θ). In this case, the opposite side is the fringe spacing (1.825 mm) and the adjacent side is the distance between the slit and the screen (3.25 m). Therefore, tan(θ) = (1.825*10^-3 m) / (3.25 m) = 5.62*10^-7. Taking the inverse tangent of this value, we get θ = 3.23*10^-7 radians.

Now, we can plug this value for θ into the formula sin(θ) = 0.334 to find the width of the slit (d). This gives us d = 1.825 mm / 0.334 = 5.46 mm. Therefore, the width of the slit is approximately 5.46 mm.

In conclusion, the mistake made in the given solution was using the formula sin(θ) = λm instead of d = λm / sin(θ). By using the correct formula and taking