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

Answer:

The ratio, $R_{\theta}$ is 2

Solution:

As per the question:

Wavelength, $\lambda = 614\ nm = 614\times 10^{- 9}\ m$

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

$\lambda = wsin\theta_{s}$

$sin\theta = \frac{\lambda}{w}$

For very small angle:

$sin\theta$ ≈ $\theta$

$\theta = \frac{\lambda}{w}$              (1)

For double slit:

n = 2

Thus

$2\lambda = wsin\theta_{s}$

$sin\theta' = \frac{\lambda}{2w}$

For very small angle:

$sin\theta$ ≈ $\theta$

$\theta' = \frac{\lambda}{2w}$             (2)

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

$R_{\theta} = \frac{\frac{\lambda}{w}}{\frac{\lambda}{2w}} = 2$