Question

In the diffraction due to a single slit experiment, the aperture of the slit is 3 mm.
If monochromatic light of wavelength 620 nm is incident normally on the slit.
calculate the separation between the first order minima and the 3rd order maxima
on one side of the screen. The distance between the slit and the screen is 1.5 m.​

Answer 1

Answer:

Distance between first order minimum and 3rd order minimum is 0.62 mm

Explanation:

As we know that the angular position of the minimum intensity on the screen is given as

$a sin\theta = N\lambda$

so we have for first order minimum

$(3 \times 10^{-3})sin\theta = 620 \times 10^{-9}$

$\theta_1 = 0.0118 degree$

now similarly for 3rd order minimum

$a\theta_2 =3\lambda$

$(3 \times 10^{-3}) sin\theta_2 = 3(620 \times 10^{-9})$

$\theta_2 = 0.0355 degree$

now the angular separation between two minimum

$\Delta \theta = 0.0355 - 0.0118$

$\Delta \theta = 0.0237 Degree$

So the distance between two minimum is

$d = L \Delta \theta$

$d = 1.5 (0.0237 \times \frac{\pi}{180})$

$d = 0.62 mm$

#Learn

Topic : Diffraction

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