Question

Light with a wavelength of λ = 614 nm is shone first on a single slit of width w = 3.75 μm. The single slit is then replaced with a double slit separated by a distance w. The ratio of the single slit angle to the double slit angle for the first dark fringe is Rθ.Find the ratio between these angles numerically.

Answer 1

The ratio,
`R_(\theta)` is 2

Solution:

As per the question:

Wavelength,
`\lambda = 614\ nm = 614* 10^(- 9)\ m`

Single slit width, w =
`3.75\ mu m`

Now,

We know from the eqn for diffraction:

`n\lambda = wsin\theta`

Now,

For single slit:

n = 1

`\lambda = wsin\theta_(s)`

`sin\theta = (\lambda)/(w)`

For very small angle:

`sin\theta` ≈
`\theta`

`\theta = (\lambda)/(w)` (1)

For double slit:

n = 2

Thus

`2\lambda = wsin\theta_(s)`

`sin\theta' = (\lambda)/(2w)`

For very small angle:

`sin\theta` ≈
`\theta`

`\theta' = (\lambda)/(2w)` (2)

For the ratio,
`R_(\theta)`, we divide en (1) by eqn (2):

`R_(\theta) = ((\lambda)/(w))/((\lambda)/(2w)) = 2`