Incomplete question.The complete question is here
If the separation between the first and the second minima of a single-slit diffraction pattern is 6.0 mm, what is the distance between the screen and the slit? The light wavelength is 500 nm and the slit width is 0.16 mm.
Answer:
The distance between first and second minimum is 1.92m
Explanation:
Given data
Wavelength of light λ=500nm
Slit width D=0.16 mm
The distance between first and second minima is 6.0 mm
To find
Distance L between the screen and the slit
Solution
As we know that
Yn=(nλL)/D
Where is L is distance between the slit and screen
D is slit width
n is order of minimum
Yn is distance of nth minimum from center maximum
λ is wavelength of light
The Distance between first and second minimum is given as:
Y₂-Y₁=6 mm
(2λL)/D-(1λL)/D=6 mm
(2λL-1λL)/D=6 mm
(λL/D)=6 mm
[tex]L=\frac{6mm(0.16mm)}{500nm}\\ L=1.92m[/tex]
The distance between first and second minimum is 1.92m
