In the diffraction pattern due to a single slit of width d with incident ligth of wavelength lamda at and angle of diffraction thita the conditon for first minimum
To find the condition for the first minimum in the diffraction pattern due to a single slit, we need to consider the interference of light waves that pass through the slit and diffract.
The condition for the first minimum occurs when the path difference between adjacent waves is exactly half a wavelength. This means that the waves interfere destructively, resulting in a dark fringe (minimum intensity).
We can use the concept of phasors to analyze this situation. Let's consider a point at a distance "x" from the center of the pattern. The path difference between the top and bottom edges of the slit can be approximated as "d*sin(θ)", where "θ" is the angle of diffraction. This path difference corresponds to a phase difference of (2π/λ) * (d*sin(θ)).
For the first minimum, the phase difference should be equal to half a wavelength, so we have:
(2π/λ) * (d*sin(θ)) = π
Simplifying the equation, we find:
d*sin(θ) = λ/2
This is the condition for the first minimum in the diffraction pattern due to a single slit. It relates the width of the slit "d", the wavelength of the incident light "λ", and the angle of diffraction "θ". To determine the exact angle, you can rearrange the equation as:
sin(θ) = λ/(2*d)
Take the inverse sine (arcsin) of both sides to find the angle of diffraction.