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

λ = a

Step-by-step 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

`(y)/(x) = (\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.