Answer:
The wide of the slit is 0.715 μm.
Explanation:
We can use the single slit diffraction equation.
[tex]d\cdot sin(\theta)=\lambda[/tex]
Where:
- d is the wide of the slit
- θ is the angle of the maximum central diffraction intensity (we use th epossitve value 35°)
- λ is the wavelength
We just need to solve this equation for d.
[tex]d=\frac{\lambda}{sin(\theta)}[/tex]
[tex]d=\frac{410*10^{-9}}{sin(35)}[/tex]
[tex]d=7.15*10^{-7}\: m[/tex]
Therefore, the wide of the slit is 0.715 μm.
I hope it helps you!
