Question

When violet light of wavelength 415 nm falls on a single slit, it creates a central diffraction peak that is 9.90 cm wide on a screen that is 2.53 m away. how wide is the slit?

Answer 1

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by

`y_n= (n \lambda D)/(a)` (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern

`\lambda` is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit
The width of the first peak (let's call it
`\Delta y`) corresponds to twice the distance of the first minimum:

`\Delta y = 2 y_1=9.90 cm=0.099 m`
which means

`y_1 = (\Delta y)/(2)= (0.099 m)/(2)=0.0495 m`

The wavelength of the light in the problem is

`\lambda=415 nm=4.15 \cdot 10^(-7) m`
and the distance of the screen from the slit is

`D=2.53 m`

So by using these data and re-arranging the equation, we can find the width of the slit:

`a= (1 \lambda D)/(y_1) = ((1)(4.15 \cdot 10^(-7) m)(2.53 m))/(0.0495 m)=2.15 \cdot 10^(-5) m = 0.0215 mm`