Question

Coherent monochromatic light of wavelength l passes through a narrow slit of width a, and a diffraction pattern is observed on a screen that is a distance x from the slit. On the screen, the width w of the central diffraction maximum is twice the distance x. What is the ratio a>l of the width of the slit to the wavelength of the light

Answer 1

Answer:

λ = a

Explanation:

This is a diffraction exercise that is described by the expression

          a sin θ = m λ

         sin θ  = m λ/ a

the first zero of the diffraction occurs for m = 1

        sin θ  = λ / a

 

angles are generally very small and are measured in radians

         sin θ  = θ  = y / x

we substitute

         $\frac{y}{x} = \frac{\lambda}{a}$

the width of the central maximum is twice the distance to zero

         w = 2y

in the exercise indicate that this width is equal to twice the distance to the screen (2x)

          W = 2x

           2y = 2x

we substitute

          1 = λ/ a

          λ = a

we see that the width of the slit is equal to the wavelength used.