Question

A narrow slit is illuminated with light of wavelength 410 nm. If the central maximum extends to ±35.0°, how wide is the slit

Answer 1

Answer:

The wide of the slit is 0.715 μm.

Explanation:  

We can use the single slit diffraction equation.

$d\cdot sin(\theta)=\lambda$

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.

$d=\frac{\lambda}{sin(\theta)}$

$d=\frac{410*10^{-9}}{sin(35)}$

$d=7.15*10^{-7}\: m$

 

Therefore, the wide of the slit is 0.715 μm.

I hope it helps you!