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Chapter 25: Problem 51

Light of wavelength 490 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen \(3.20 \mathrm{m}\) from the slit. The distance on the screen between the central maximum and the third minimum is $2.5 \mathrm{cm} .$ What is the width of the slit?

Short Answer

Expert verified
Answer: The width of the slit is approximately 3.22 碌m.

Step by step solution

01

Identify the given information

The given information in the problem is as follows: - Wavelength of light (\(\lambda\)): \(490\,\mathrm{nm}\) - Distance between the slit and the screen (L): \(3.20\,\mathrm{m}\) - Distance on the screen between the central maximum and the third minimum (y): \(2.5\,\mathrm{cm}\) Keep in mind that we need to convert the given distances to the same unit (meters).
02

Convert the distances to meters

Convert the distances to meters using the following conversions: - \(\lambda = 490\,\mathrm{nm} \times 10^{-9}\,\frac{\mathrm{m}}{\mathrm{nm}} = 490 \times 10^{-9}\,\mathrm{m}\) - \(y = 2.5\,\mathrm{cm} \times 10^{-2}\,\frac{\mathrm{m}}{\mathrm{cm}} = 2.5 \times 10^{-2}\,\mathrm{m}\) Now, we have \(\lambda = 490 \times 10^{-9}\,\mathrm{m}\), L = \(3.20\,\mathrm{m}\), and y = \(2.5 \times 10^{-2}\,\mathrm{m}\).
03

Use the formula for the angular position of the minima

The formula for the angular position of the minima in a single-slit diffraction pattern is given by: $$ m\lambda = a\sin\theta $$ where: - m is the order of the minimum (m = 3 for the third minimum) - a is the width of the slit - \(\theta\) is the angle between the central maximum and the minimum First, we need to find the angle \(\theta\).
04

Calculate the angle 胃

Use the geometry of the situation to calculate the angle \(\theta\). In this case, the slit is the adjacent side, y is the opposite side of the right triangle, and L is the hypotenuse. Therefore, we have: $$ \tan\theta = \frac{y}{L} \Rightarrow \theta = \arctan{\frac{y}{L}} $$ Plugging in the values for y and L, we get: $$ \theta = \arctan{\frac{2.5 \times 10^{-2}\,\mathrm{m}}{3.20\,\mathrm{m}}} \approx 0.449\,\mathrm{rad} $$
05

Substitute the values into the formula for minima and solve for a

Now we are ready to substitute the values into the formula to find the width of the slit (a). For the third minimum, m = 3. We have: $$ 3(490 \times 10^{-9}\,\mathrm{m}) = a\sin{0.449\,\mathrm{rad}} $$ Solve for a: $$ a = \frac{3(490 \times 10^{-9}\,\mathrm{m})}{\sin{0.449\,\mathrm{rad}}} \approx 3.22 \times 10^{-6}\,\mathrm{m} $$
06

Convert the width of the slit to micrometers and present the final answer

Convert the width of the slit from meters to micrometers: $$ a = 3.22 \times 10^{-6}\,\mathrm{m} \times 10^{6}\,\frac{\mathrm{\mu m}}{\mathrm{m}} \approx 3.22\,\mathrm{\mu m} $$ Hence, the width of the slit is approximately \(3.22\,\mathrm{\mu m}\).

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Most popular questions from this chapter

Two radio towers are a distance \(d\) apart as shown in the overhead view. Each antenna by itself would radiate equally in all directions in a horizontal plane. The radio waves have the same frequency and start out in phase. A detector is moved in a circle around the towers at a distance of $100 \mathrm{km}.$ The waves have frequency \(3.0 \mathrm{MHz}\) and the distance between antennas is \(d=0.30 \mathrm{km} .\) (a) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=0^{\circ} ?\) (b) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=90^{\circ} ?\) (c) At how many angles $\left(0 \leq \theta<360^{\circ}\right)$ would you expect to detect a maximum intensity? Explain. (d) Find the angles \((\theta)\) of the maxima in the first quadrant \(\left(0 \leq \theta \leq 90^{\circ}\right) .\) (e) Which (if any) of your answers to parts (a) to (d) would change if the detector were instead only $1 \mathrm{km}$ from the towers? Explain. (Don't calculate new values for the answers.)

Young's Double-Slit Experiment Light of \(650 \mathrm{nm}\) is incident on two slits. A maximum is seen at an angle of \(4.10^{\circ}\) and a minimum of \(4.78^{\circ} .\) What is the order \(m\) of the maximum and what is the distance \(d\) between the slits?
Light incident on a pair of slits produces an interference pattern on a screen \(2.50 \mathrm{m}\) from the slits. If the slit separation is $0.0150 \mathrm{cm}$ and the distance between adjacent bright fringes in the pattern is \(0.760 \mathrm{cm},\) what is the wavelength of the light? [Hint: Is the small-angle approximation justified?]
A camera lens \((n=1.50)\) is coated with a thin film of magnesium fluoride \((n=1.38)\) of thickness \(90.0 \mathrm{nm}\) What wavelength in the visible spectrum is most strongly transmitted through the film?
A grating \(1.600 \mathrm{cm}\) wide has exactly 12000 slits. The grating is used to resolve two nearly equal wavelengths in a light source: \(\lambda_{\mathrm{a}}=440.000 \mathrm{nm}\) and $\lambda_{\mathrm{b}}=440.936 \mathrm{nm}$ (a) How many orders of the lines can be seen with the grating? (b) What is the angular separation \(\theta_{b}-\theta_{a}\) between the lines in each order? (c) Which order best resolves the two lines? Explain.
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