How Does N-Slit Interference Affect Minima Spacing?

[shaiqbashir]

Help Required urgently in this light problem

Hi !

can anybody help me in solving this problem!


The electric field at Poin P due to N slits is,


A(r,\theta,t) =A(r) \frac{\sin\frac{1}{2}\N\Delta\phi\cos(kr-\omega\ t)}{\sin\frac{1}{2}\Delta\phi}

where \Delta\phi =kd\sin\theta,

"d" being the spacing between adjacent slits.Let N=4.

a) Find two anglws on either side of \theta=0 at which there is an interference minima.
b)Find the spacing between the minima compare tp the case pf only two slits of separation "d".

I have solved its "a" part but i just couldn't solve its "b" part. Plz help mw in solving this:

Thanks a lot!

 

[member 553083]

Solution: a) For interference minima, A(r,\theta,t) =0\frac{\sin\frac{1}{2}\N\Delta\phi\cos(kr-\omega\ t)}{\sin\frac{1}{2}\Delta\phi} = 0 \frac{1}{2}\N\Delta\phi\cos(kr-\omega\ t) = n\pi where, n = 0,1,2,3.....\Delta\phi = \frac{2n\pi}{N\cos(kr-\omega\ t)} \Delta\phi = \frac{8n\pi}{4\cos(kr-\omega\ t)} \Delta\phi = 2n\pi\tan\theta \therefore \theta = \frac{n\pi}{2}\cot^{-1} kd For N=4, \theta = \frac{n\pi}{4}\cot^{-1} kd Where n = 0,1,2,3Therefore, the angles on either side of \theta=0 at which there is an interference minima are \theta = \pm \frac{\pi}{4}\cot^{-1} kd b) The spacing between the minima for four slits is D_4 = 2\theta = \pm \frac{\pi}{2}\cot^{-1} kd The spacing between the minima for two slits isD_2 = \frac{\pi}{\cot^{-1} kd} Therefore, the ratio of the spacing between minima of four slits to that of two slits is \frac{D_4}{D_2} = \frac{\pm \frac{\pi}{2}\cot^{-1} kd}{\frac{\pi}{\cot^{-1} kd}} = \pm \frac{1

 

[thepatientmental]

Hi there,

I can definitely try to help you with this problem! It seems like you have already made some progress with part a) of the question, which is great. For part b), we can use the formula for the spacing between minima, which is given by:

\Delta\theta = \frac{\lambda}{d(N-1)}

Where \lambda is the wavelength of light. Since we are given the number of slits (N=4) and the spacing between them (d), we just need to find the wavelength of light to solve for the spacing between minima. To do this, we can use the formula \lambda=\frac{2\pi}{k}, where k is the wavenumber. The wavenumber can be found using the relation k=\frac{2\pi}{\lambda}=2\pi\nu, where \nu is the frequency of light. Thus, we can calculate the wavelength using the frequency of light and then use it to find the spacing between minima using the formula above.

I hope this helps! Let me know if you need any further assistance. Good luck with your problem!