Diffraction Grating and Intensity

[silence98]

Homework Statement

Show that the maximum intensity for a diffraction grating of N slits at the first principal maximum is N^2 times bigger than for a single slit.

Homework Equations

I=I(0)[sin^2(NB)/sin^2(B)] where B is ($$\pi$$dsin$$\phi$$)/$$\lambda$$

The Attempt at a Solution

I feel i understand the concepts here but this question has stumped me. I've scoured the internet, read through the relevant chapters in my two textbooks and still i can't see how this would be proven. It is just stated in my lecture notes, is it that intuitive?

I really would be grateful for any help.

 

[ehild]

The maximum is at Φ=0. You have to start with the limit sin(x)/x when x tends to 0. Find the limit of sin(Nx)/sin(x). Use that sin(Nx)/sin(x)=[sin(Nx)/Nx] N [x/sin(x)].

ehild

 

[silence98]

The maximum is at Φ=0. You have to start with the limit sin(x)/x when x tends to 0. Find the limit of sin(Nx)/sin(x). Use that sin(Nx)/sin(x)=[sin(Nx)/Nx] N [x/sin(x)].

ehild

Thanks for the help, but i don't really understand the bolded part. I haven't seen that trig identity before and i can't see how it would be found.

 

[ehild]

It is just multiplying both the numerator and the denominator with the same quantity and rearranging.

$$\frac{\sin(Nx)}{\sin(x)}=\frac{Nx\sin(Nx)}{Nx\sin(x)}=\frac{\sin(Nx)}{Nx} N \frac{x}{\sin(x)}$$

ehild

 

[silence98]

It is just multiplying both the numerator and the denominator with the same quantity and rearranging.

$$\frac{\sin(Nx)}{\sin(x)}=\frac{Nx\sin(Nx)}{Nx\sin(x)}=\frac{\sin(Nx)}{Nx} N \frac{x}{\sin(x)}$$

ehild

Thankyou, and i apologise for my stupidity!

I've got it now.