Wave Optics - Single Slit Diffraction

[Alice-Shallom]

Light of wavelength 600nm falls on a 0.40-mm-wide slit and forms a diffraction pattern on a screen 1.5m away.
(a) Find the position of the first dark band on each side of the central maximum.
(b) Find the width of the central maximum


For the (a) I think that we use sinθ = +- λ / α
I think m is equal to : m = +- 1

Where λ = 600 nm
and α = 0.40 mm


If the way I have solved (a) is correct, then how do I proceed to (b) ? I know L is 1.5m then what?

 

[turin]

You use trig. That θ in the equation is the angle for a particular right triangle. There should be a picture in your book. L is one leg and the width of the central max is twice the other leg.

Actually, that equation you used for (a) may need an adjustment. I'll get back to you.

I think that destructive interference occurs for the condition:

(α/2) sinθ = λ/2

Oh, sorry. You just manipulated it. The way you have it should be correct.

 

[M98Ranger]

For part (a), your approach is correct. To find the position of the first dark band on each side of the central maximum, you can use the equation sinθ = ± mλ/α, where m is the order of the dark band (in this case, m = ±1), λ is the wavelength of light (600 nm), and α is the width of the slit (0.40 mm).

So for the first dark band on the left side, we have sinθ = -1*(600nm)/(0.40mm) = -0.0015. Taking the inverse sine of this value, we get θ = -0.086 degrees. Similarly, for the first dark band on the right side, we have sinθ = 1*(600nm)/(0.40mm) = 0.0015, giving θ = 0.086 degrees.

For part (b), we can use the equation w = 2λL/α, where w is the width of the central maximum, λ is the wavelength of light, L is the distance from the slit to the screen, and α is the width of the slit. Plugging in the values, we get w = 2*(600nm)*(1.5m)/(0.40mm) = 4.5 mm. So the width of the central maximum is 4.5 mm.

 

[M98Ranger]

Your approach for part (a) is correct. To find the position of the first dark band on each side of the central maximum, we can use the equation sinθ = ± mλ/α, where m is the order of the dark band, λ is the wavelength of light, and α is the width of the slit. In this case, m = ±1, λ = 600 nm, and α = 0.40 mm. Plugging these values into the equation, we get:

sinθ = ± (1)(600 nm) / (0.40 mm) = ± 1.5

To find the position of the first dark band, we need to solve for θ. Taking the inverse sine of both sides, we get:

θ = sin^-1(± 1.5) = ± 54.6°

Since there are two first dark bands on each side of the central maximum, we have two values for θ: ±54.6°. To find the actual position on the screen, we can use the small angle approximation formula: θ = y/L, where y is the distance from the central maximum to the first dark band and L is the distance from the slit to the screen. Plugging in the values, we get:

y = (±54.6°)(1.5 m) = ± 0.82 m

Therefore, the first dark bands on each side of the central maximum are located at ±0.82 m from the central maximum.

For part (b), we can use the equation for the width of the central maximum, which is given by:

w = 2λL/α

Plugging in the values, we get:

w = (2)(600 nm)(1.5 m) / (0.40 mm) = 4.5 mm

Therefore, the width of the central maximum is 4.5 mm. This means that the central maximum will cover an area of 4.5 mm on the screen, while the first dark bands will appear 0.82 m away from the central maximum on each side.