Answer:
The slit width is 2.31 μm
Explanation:
Given that
Angle θ = 15°
Wavelength λ = 600 nm
We know that relationship between slit width and wavelength
[tex]w=\dfrac{\lambda }{sin\theta }[/tex]
Now by putting the values
[tex]w=\dfrac{\lambda }{sin\theta }[/tex]
[tex]w=\dfrac{600 }{sin15^{\circ} }[/tex]
w=2318.22 nm
w= 2.31 μm
So the slit width is 2.31 μm
