Question

Light with wavelength 633 nm is incident on a 3.00-μm-wide slit.Part AFind the angular width of the central peak in the diffraction pattern, taken as the angular separation between the first minima.

Answer 1

Answer:

The angular width of the central peak is 24.2 degrees

Explanation:

The equation that describes the single-slit diffraction phenomenon is:

$a\sin\theta=m\lambda$ (1)

with a the width of the slit, $\theta$ the angular position of the minimum regarding the center of the screen where light is projected, m the order of the minimum and $\lambda$ the wavelength. Solving (1) for $\theta$ with m=1 that is the first minimum:

$\theta=\arcsin(\frac{m\lambda}{a})=\arcsin(\frac{(1)(633\times10^{-9})}{3.0\times10^{-6}})\approx12.1 deg$

See the figure below that the central peak is symmetric regarding the center of the screen, which implies that the angular width of the central peak is $2\theta =2(12.1 deg) = 24.2 deg$