Question

An experiment is set-up with a monochromatic light source, a double slit, and a screen, similar to what is shown on page 7-2 of your lab manual. The experimenter measures and sketches the intensity pattern shown below. The scale along the bottom of the figure is in centimeters. What is the distance 2x3 between the third maxima in the intensity pattern? Note: the central peak is the ?zeroth maximum?, the two peaks closest to this on either side are the ?first maxima?, the next two peaks are the ?second maxima?, etc. (Units required, and you must be within 1 mm to be correct.) The slit-to-screen distance is D = 2.000 m, and the slit separation is 0.186 mm. Use the formula for double-slit diffraction maxima and your value of 2x3 to find the light source?s wavelength lambda. (Units required.) lambda = If the light source?s wavelength were set to 483 nm, but everything else left the same, what would the distance between the second maxima 2x2 become? (Units required.) When lambda = 483 nm then 2x2 =

Answer 1

The picture is not so clear, as far as I can see

\(2 x_{3}=38 \mathrm{~mm}\)

In Young's double slit experiment

\(y=\frac{m \lambda D}{d}\)

\(\lambda=\frac{y d}{m D}\)

\(\lambda=\frac{0.186 \times 19}{3 \times 2000}\)

\(=589 \mathrm{nm}\)

Wavelength \(589 \mathrm{nm}\)

\(y=\frac{m \lambda D}{d}\)

\(=\frac{2 \times 483 \times 10^{-6} \times 2000}{0.186}\)

\(=10.38 \mathrm{~mm}\)

The maxima would be \(20.76 \mathrm{~mm}\)