Monochromatic light of wavelength $441 n m$ is incident on a narrow slit. On a screen $2.00 m$ away, the distance between the second diffraction minimum and the central maximum is $1.50 c m$ . Find the width of the slit.

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Solution

For the $m$ th diffraction minimum, a $sin θ = m λ$ .
We solve for the slit width:
$a = \frac{m λ}{sin θ} = \frac{2 (441 n m)}{sin {0.430}^{\circ}} = 0.118 m m$

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Monochromatic light of wavelength 441nm is incident on a narrow slit. On a screen 2.00m away, the distance between the second diffraction minimum and the central maximum is 1.50cm. Find the width of the slit.

For the m th diffraction minimum, a sinθ=mλ. We solve for the slit width:a=mλsinθ=2(441nm)sin0.430∘=0.118mm

Monochromatic light of wavelength $441 n m$ is incident on a narrow slit. On a screen $2.00 m$ away, the distance between the second diffraction minimum and the central maximum is $1.50 c m$ . Find the width of the slit.

For the $m$ th diffraction minimum, a $sin θ = m λ$ .
We solve for the slit width:
$a = \frac{m λ}{sin θ} = \frac{2 (441 n m)}{sin {0.430}^{\circ}} = 0.118 m m$