Does Widening the Slit Increase Brightness in Young's Slit Experiment?

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Widening the slit in Young's slit experiment increases the intensity of light at the maxima due to a higher concentration of photons reaching the central maxima. While less diffraction occurs with a wider slit, leading to a narrower spread of light, the overall irradiance increases. The discussion clarifies that intensity and irradiance are distinct, with intensity measured in W/Steradian and irradiance in W/m^2. The relationship between slit width and diffraction is nuanced, as less diffraction can lead to greater intensity despite potential decreases in interference effects. Ultimately, a wider slit results in brighter maxima due to increased photon concentration.
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Young's slit experiment question!Help me

On a young's slit experiment,i found that,even at the maxima,the wave intensity was small.If i had made the slit wider to let more energy through,would the wave intensity be brighter at maxima?If not why?
 
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Yes, the irradiance (and the intensity) would be greater because the light is diffracting over a smaller angle.

Note that the units for intensity are W/Steradian, the units for irradiance are W/m^2. I suspect you are actually referring to irradiance.

Claude.
 
I think if i made the slit wider,

>that means the diffraction is less
>less diffraction=less interference
>which means the intensity (brightneess)decreases.

Correct me if i am wrong please!
 
You are sort of correct by say "diffraction is less," if you mean that the beam is less spread out. If the slit is widened, the path of the photons would become less uncertain, meaning the photon concentration that falls on the central maxima will be greater. This means greater intensity.
 
What do you mean by 'less interference'?

Claude.
 
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Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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