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Chapter 32: Problem 12

A double-slit experiment has slit spacing \(0.12 \mathrm{mm}\). (a) What should be the slit-to-screen distance \(L\) if the bright fringes are to be \(5.0 \mathrm{mm}\) apart when the slits are illuminated with 633 -nm laser light? (b) What will be the fringe spacing with 480 -nm light?

Short Answer

Expert verified
The required slit-to-screen distance should be approximately 0.95 m to produce a fringe spacing of 5.0 mm with 633-nm light. If a 480-nm light is used instead, the new fringe spacing will approximately be 3.8 mm.

Step by step solution

01

Calculate the slit-to-screen distance

Rewrite the formula as \(L=\dfrac{y \cdot d}{\lambda}\) and substitute the given values into the formula to determine the slit-to-screen distance. Here, \(y = 5.0 \mathrm{mm}\) is the fringe spacing, \(d = 0.12 \mathrm{mm}\) is the slit spacing and \(λ = 633 \mathrm{nm}\) is the wavelength of the light.
02

Calculate the fringe spacing for the different wavelength

Substitute \(L\) from step 1, \(\lambda = 480 \mathrm{nm}\) as the new wavelength and the given slit spacing \(d = 0.12 \mathrm{mm}\) into the formula \(y=\dfrac{L \cdot \lambda}{d}\) to find the new fringe spacing.

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