Problem involving slits, diffraction, and incident angle.

[nobb]

Hi.
I am a bit confused with this question:

A plane wave of 400-nm light is incident on a 25-µ(mu) m slit in a screen, as shown in the figure below. At what incident angle will the first null of the diffraction pattern be on a line perpendicular to the screen? (picture is attached).

Do I use the formula wavelength=(dSinA)/n ? I am a bit unsure what the "first null" refers to and how this should be taken into account into my formula. Is the first null referring to "n"?

Help would be appreciated. Thank you.

 

[andrevdh]

I would think that the first null will refer to the position next to the central maximum where destructive interference occurs "for the first time". The formula you are referring to gives the angles at which minima will be located behind the screen. So the first minima will occur at n = 1.

 

[kyle8921]

I can help clarify this question for you. The first null in a diffraction pattern refers to the first point where there is no detectable light intensity. In other words, it is where the light waves cancel each other out and there is no resulting light intensity on the screen. This is commonly referred to as the dark fringe or the first minimum in the diffraction pattern.

To calculate the incident angle at which the first null will occur, you can use the equation you mentioned: wavelength = (d*sin θ)/n. In this equation, d represents the slit width, θ represents the incident angle, and n represents the order of the diffraction pattern (in this case, n=1 for the first null).

So, to find the incident angle, you can rearrange the equation to solve for θ: θ = arcsin (n*wavelength/d). Plugging in the given values (n=1, wavelength=400 nm, d=25 µm), we get θ = arcsin (1*400 nm / 25 µm) = 9.6 degrees. This means that the first null will occur at an incident angle of 9.6 degrees.

I hope this helps clarify the question for you. If you have any further questions, please don't hesitate to ask.