Intensity from a multi slit grating

[UrbanXrisis]

The intensity from a multi slit grating with 2N+1 slites is given by:

$$I(Y)=\frac{M^2}{2R^2} \frac{sin^2 \left[(2N+1) \frac{x}{2}\right]}{sin^2 \left[\frac{x}{2}\right]}$$

where $$x=\frac{2 \pi Y d}{D \lambda}$$

I am to show how to find the limiting case of just two slites and show that in this limit the formula is exactly the same as:

$$I(Y)=\frac{2M^2}{D^2} cos^2 \left(\frac {\pi d sin\theta}{\lambda}\right)$$

so what I did:

$$I(Y)=\frac{M^2}{2R^2} \frac{sin^2 \left[(2N+1) \frac{x}{2}\right]}{sin^2 \left[\frac{x}{2}\right]}$$

$$=\frac{M^2}{2R^2} \frac{sin^2 \left[ \frac{2 \pi \lambda d}{D \lambda} \right]}{sin^2 \left[\frac{\pi \lambda d}{D \lambda} \right]}$$


I could turn this into cos/cos but I don't know how to get it into cos(sin) any ideas?

 

[Nate810]

The answer is actually:I(Y)=\frac{M^2}{2R^2} \frac{sin^2 \left[(2N+1) \frac{x}{2}\right]}{sin^2 \left[\frac{x}{2}\right]}=\frac{M^2}{2R^2} \frac{sin^2 \left[ \frac{2 \pi \lambda d}{D \lambda} \right]}{sin^2 \left[\frac{\pi \lambda d}{D \lambda} \right]}=\frac{M^2}{2R^2} \frac{cos^2 \left[ \frac{\pi \lambda d}{D \lambda} \right]}{sin^2 \left[\frac{\pi \lambda d}{D \lambda} \right]}=\frac{M^2}{2R^2} cos^2 \left(\frac {\pi d sin\theta}{\lambda}\right)=\frac{2M^2}{D^2} cos^2 \left(\frac {\pi d sin\theta}{\lambda}\right)

 

[quicknote]

Firstly, it is important to note that the formula given is for the intensity at a particular point Y, which represents the distance from the central maximum of the diffraction pattern. This means that the formula is dependent on the distance between the slits (d), the wavelength of the light (λ), and the distance between the grating and the screen (D). M represents the overall brightness of the pattern, and R represents the distance between the grating and the observer's eye.

To find the limiting case of just two slits, we can substitute N=1 into the formula:

I(Y)=\frac{M^2}{2R^2} \frac{sin^2 \left[(2(1)+1) \frac{x}{2}\right]}{sin^2 \left[\frac{x}{2}\right]} = \frac{M^2}{2R^2} \frac{sin^2 \left[\frac{3x}{2}\right]}{sin^2 \left[\frac{x}{2}\right]}

We can also rewrite the value of x as:

x = \frac{2 \pi Y d}{D \lambda} = \frac{2 \pi d sin\theta}{\lambda}

where θ represents the angle of diffraction.

Substituting this into the formula, we get:

I(Y)=\frac{M^2}{2R^2} \frac{sin^2 \left[\frac{3}{2} \left( \frac{2 \pi d sin\theta}{\lambda} \right) \right]}{sin^2 \left[\frac{1}{2} \left( \frac{2 \pi d sin\theta}{\lambda} \right) \right]}

Using the trigonometric identity cos(2θ) = 2cos^2(θ) - 1, we can rewrite this as:

I(Y)=\frac{M^2}{2R^2} \frac{1}{sin^2 \left[\frac{1}{2} \left( \frac{2 \pi d sin\theta}{\lambda} \right) \right]} \left[ 1 - cos^2 \left( \frac{2 \pi d sin\theta}{\lambda} \right) \right]

Now, if we take the limit as N approaches infinity, the term