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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.

Respuesta :

Answer:

The ratio, [tex]R_{\theta}[/tex] is 2

Solution:

As per the question:

Wavelength, [tex]\lambda = 614\ nm = 614\times 10^{- 9}\ m[/tex]

Single slit width, w = [tex]3.75\ mu m[/tex]

Now,

We know from the eqn for diffraction:

[tex]n\lambda = wsin\theta[/tex]

Now,

For single slit:

n = 1

[tex]\lambda = wsin\theta_{s}[/tex]

[tex]sin\theta = \frac{\lambda}{w}[/tex]

For very small angle:

[tex]sin\theta[/tex] ≈ [tex]\theta[/tex]

[tex]\theta = \frac{\lambda}{w}[/tex]              (1)

For double slit:

n = 2

Thus

[tex]2\lambda = wsin\theta_{s}[/tex]

[tex]sin\theta' = \frac{\lambda}{2w}[/tex]

For very small angle:

[tex]sin\theta[/tex] ≈ [tex]\theta[/tex]

[tex]\theta' = \frac{\lambda}{2w}[/tex]             (2)

For the ratio, [tex]R_{\theta}[/tex], we divide en (1) by eqn (2):

[tex]R_{\theta} = \frac{\frac{\lambda}{w}}{\frac{\lambda}{2w}} = 2[/tex]