Question

Monochromatic light of wavelength 441 $\mathrm{nm}$ is incident on a narrow slit. On a screen 2.00 $\mathrm{m}$ away, the distance between the second diffraction minimum and the central maximum is 1.50 $\mathrm{cm} .$ (a) Calculate the angle of diffraction $\theta$ of the second minimum. (b) Find the width of the slit.

          Monochromatic light of wavelength 441 $\mathrm{nm}$ is incident on a
narrow slit. On a screen 2.00 $\mathrm{m}$ away, the distance between the
second diffraction minimum and the central maximum is 1.50 $\mathrm{cm} .$
(a) Calculate the angle of diffraction $\theta$ of the second minimum.
(b) Find the width of the slit.

Added by Todd L.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Ameer Said

11/30/2021

Step 1

- Wavelength of light, $\lambda = 441 \, \mathrm{nm} = 441 \times 10^{-9} \, \mathrm{m}$ - Distance from the slit to the screen, $L = 2.00 \, \mathrm{m}$ - Distance between the second diffraction minimum and the central maximum, \(y = 1.50 \, \mathrm{cm} = Show more…

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Monochromatic light of wavelength 441 $\mathrm{nm}$ is incident on a narrow slit. On a screen 2.00 $\mathrm{m}$ away, the distance between the second diffraction minimum and the central maximum is 1.50 $\mathrm{cm} .$ (a) Calculate the angle of diffraction $\theta$ of the second minimum. (b) Find the width of the slit.