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

Sagot :

whitneytr12

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.

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