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

The angular width of the central peak is 24.2 degrees

Step-by-step 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((m\lambda)/(a))=\arcsin(((1)(633*10^(-9)))/(3.0*10^(-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`