Answer: Diffraction Maxima: Angular Locations of 1-4th Maxima

[shyta]

Homework Statement

Show that the angular locations of the first to fourth secondary maximas are $$\alpha$$ = a sin $$\Theta$$/$$\Lambda$$ = 1.43030 2.45902 3.47089 4.46641 respectively.

a is the slit width = 0.00016m
$$\Lambda$$ wavelength of laser 650nm

Homework Equations

I(_$$\Theta$$) = I₀ [sin($$\Pi\alpha$$)/$$\Pi\alpha$$]²

where I₀ is the intensity at the central peak

The Attempt at a Solution

I have no idea how to begin..

 

[sungod]

You have an equation for the intensity of the diffraction pattern. The maxima and minima of the function will give the maxima and minima of the diffraction pattern. So, simply differentiate and set to zero. You'll get two factors. One of these gives the minima and the other gives the maxima, or, at least, another equation for the maxima. Solving the "maxima equation" will give the values at which the maxima occur.

 

[shyta]

hi there
thanks for your reply,

after differentiating, i got this equation

2I₀ sin(_{$$\pi$$$$\alpha$$})/$$\pi$$^2$$\alpha$$^3 ($$\pi \alpha$$ cos($$\pi\alpha$$) - sin $$\pi\alpha$$ )

i hope this is correct.. anyway i set this to zero and got 2 different sets of eqn.2I₀ sin($$\pi$$$$\alpha$$)/$$\pi$$^2$$\alpha$$^3 = 0

so sin($$\pi$$$$\alpha$$) = 0
$$\alpha$$ = 0,1,2,3,...

and ($$\alpha$$^2 cos($$\pi\alpha$$) - sin $$\pi\alpha$$ ) = 0
$$\pi\alpha$$ = tan ($$\pi\alpha$$)
how do I solve this?

 

[shyta]

sorry the correct maxima eqn should be this.

$$\pi$$$$\alpha$$ = tan ($$\pi$$$$\alpha$$)

how do i solve it?

 

[sungod]

Let me rewrite this as
b = tan(b)

Plot y = b and y = tan(b) on the same plot. Where the two plots intersect are the solutions.