Diffraction minima:

$\overline{){\mathbf{a}}{\mathbf{}}{\mathbf{s}}{\mathbf{i}}{\mathbf{n}}{\mathbf{\theta}}{\mathbf{=}}{\mathbf{m}}{\mathbf{\lambda}}}$

Wavelength in water:

$\overline{){{\mathbf{\lambda}}}_{{\mathbf{w}}}{\mathbf{=}}\frac{{\mathbf{\lambda}}_{\mathbf{a}\mathbf{i}\mathbf{r}}}{{\mathbf{\eta}}_{\mathbf{w}}}}$

**Part A**

For small angles measured in radians;

tan θ ≈ sin θ ≈ opposite/adjacent

opposite = y = (17.9 mm)/2 = 8.95 mm (1m/1000mm) = 8.95 × 10^{-3} m

adjacent = D = 80.0 cm(1m/100cm) = 0.8 m

λ = 633 nm (10^{-9}m/1nm) = 633 × 10^{-9} m

You have been asked to measure the width of a slit in a piece of paper. You mount the paper 80.0 centimeters from a screen and illuminate it from behind with laser light of wavelength 633 nanometers (in air). You mark two of the intensity minima as shown in the figure, and measure the distance between them to be 17.9millimeters. (Figure 1)

Part A

What is the width, a, of the slit?

Express your answer in micrometers, to three significant figures.

Part B

If the entire apparatus were submerged in water, would the width of the central peak change?

a) The width would increase.

b) The width would decrease.

c) The width would not change.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Single Slit Diffraction concept. You can view video lessons to learn Single Slit Diffraction. Or if you need more Single Slit Diffraction practice, you can also practice Single Slit Diffraction practice problems.

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Based on our data, we think this problem is relevant for Professor Holczer's class at UCLA.