N-slit Interference/Diffraction

[Warr]

Hi, having trouble determining whether my lab book has made an error, or I have.

Intensity as a function of $$\phi$$ for 2 slits is given as

$$A^2={A_2}^2\frac{sin^2(\frac{\phi}{2})}{{(\frac{\phi}{2})}^2}cos^2(\frac{\beta}{2})$$

but then it gives the amplitude for N slits to be

$$A^2={A_N}^2\frac{sin^2(\frac{\phi}{2})}{{(\frac{\phi}{2})}^2}\frac{sin^2(\frac{N\beta}{2})}{sin^2(\frac{\beta}{2})}$$

However, when I sub in N = 2 for the equation (2), and use the double angle formula to reduce the right fraction in equation (2) I get 4*equation(1) rather than just the equation(1) alone. Am I doing it wrong?

To be more succinct, isn't $$\frac{sin^2(\frac{N\beta}{2})}{sin^2(\frac{\beta}{2})} = 4cos^2(\frac{\beta}{2})$$?

 

[Doc Al]

I don't have my optics references handy, but I don't think you're doing anything wrong.

but then it gives the amplitude for N slits to be

$$A^2={A_N}^2\frac{sin^2(\frac{\phi}{2})}{{(\frac{\phi}{2})}^2}\frac{sin^2(\frac{N\beta}{2})}{sin^2(\frac{\beta}{2})}$$

I'd say that should be:
$$A^2={A_0}^2\frac{sin^2(\frac{\phi}{2})}{{(\frac{\phi}{2})}^2}\frac{sin^2(\frac{N\beta}{2})}{sin^2(\frac{\beta}{2})}$$

It makes sense that the peak intensity should be proportional to the number of slits squared.

 

[kyle8921]

Hello, it appears that there may be a mistake in your lab book. The equation for N slits should be A^2 = A_N^2 * (sin^2(phi/2) / (phi/2)^2) * (sin^2(N*beta/2) / sin^2(beta/2)). This is different from the equation you have written, which seems to be missing the first fraction. When substituting N=2, the equation should simplify to 4*equation (1), as you have correctly pointed out. I would recommend double checking your lab book and possibly consulting with your instructor to clarify any confusion.