What happens to the diffraction pattern when you have a small wavelength and large slit width in single slit interference?
Say you have a light of 700nm shining through a diffraction grating with the separation of slits being 450nm. What pattern would you see?
The answer is given is a single, sharp peak.
I'm confused by this answer.
I know that d * sin(theta) = m * lambda
since theta = arcsin (m*lambda/d)
theta = arcsin(1.5556), which is invalid. So what is the reason for the sharp peak?
In single slit interference, the diffraction pattern is essentially a series of bright and dark regions that result from the interference of light waves passing through a single narrow slit. The exact pattern depends on the wavelength of light and the width of the slit.
When the slit width is large compared to the wavelength of light, the diffraction pattern becomes less pronounced, and the main feature is a single, sharp central peak. This occurs because the light passing through the slit spreads out less, resulting in less interference between the waves.
In your example, the light has a wavelength of 700 nm and is passing through a diffraction grating with a slit separation of 450 nm. The slit separation is smaller than the wavelength, which means the slit width is relatively large compared to the wavelength. As a result, the interference effects are minimized, and you would observe a single, sharp peak in the diffraction pattern. This peak corresponds to the central maximum of the interference pattern.
The reason why you are getting an invalid angle according to the formula you mentioned, θ = arcsin(m*λ/d), is that the value of sine of the angle arcsin(1.5556) is not within the range -1 to 1. This suggests that the angle you are calculating does not correspond to a valid diffraction order for the given parameters. However, it does not affect the observation of the single, sharp peak in the diffraction pattern.