# I Single slit diffraction

1. Dec 16, 2017

### Das apashanka

In a single slit experiment if the condition of interference maxima be asinΘ=nλ where n=1,2,3,4....
and the condition of diffraction minima is also the same
Will it not cause any effect on the intensity of the interference maxima as both are at a same distance from the central maximum on the screen?
a is the length of the slit

2. Dec 16, 2017

### lekh2003

Is your question referring to the decreasing intensity of the interference maxima as the distance increases from the center? Could you please be more specific or clear?

By the way, the interference and destruction maxima and minima have different equations for single slit interference. You can find the minima from m, but the maxima from [m+1/2].

3. Dec 16, 2017

### Das apashanka

Actually I want to clarify that for a single slit experiment if ,a be the slit width
Then what will be the condition of interference minima
Is it asinΘ=(2n-1)λ/2 ,n=1,2,3 4.....taking path difference between rays from the two edges
Or
asinΘ=(2n-1)λ,n=1,2,3,4.....taking path difference between two rays from the centre and from one edge

4. Dec 16, 2017

### lekh2003

If a is the slit width, θ is the angle of the minima, λ is the wavelength of the light, and n is a value, then this equation will apply for the destruction minima:

a sinθ = nλ

For interference maxima, the following equation applies:

a sinθ = n(λ+1/2)

5. Dec 16, 2017

### Mister T

That's the correct expression for diffraction minima where $a$ is the slit width.

The expression for double-slit interference maxima is $d \sin \theta = n \lambda$ where $d$ is the slit separation distance (and $n$ can also equal $0$ for the central maximum).