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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 µm.

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 θ

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

Sketch a sinusoidal wave with an amplitude of \(2 \mathrm{cm}\) and a wavelength of \(6 \mathrm{cm} .\) This wave represents the electric field portion of a visible EM wave traveling to the right with intensity \(I_{0}\). (a) Sketch an identical wave beneath the first. What is the amplitude (in centimeters) of the sum of these waves? (b) What is the intensity of the new wave? (c) Sketch two more coherent waves beneath the others, one of amplitude \(3 \mathrm{cm}\) and one of amplitude \(1 \mathrm{cm},\) so all four are in phase. What is the amplitude of the four waves added together? (d) What intensity results from adding the four waves?
Parallel light of wavelength \(\lambda\) strikes a slit of width \(a\) at normal incidence. The light is viewed on a screen that is \(1.0 \mathrm{m}\) past the slits. In each case that follows, sketch the intensity on the screen as a function of \(x\), the distance from the center of the screen, for $0 \leq x \leq 10 \mathrm{cm}$ (a) \(\lambda=10 a\). (b) \(10 \lambda=a,\) (c) \(30 \lambda=a.\)
Thin Films At a science museum, Marlow looks down into a display case and sees two pieces of very flat glass lying on top of each other with light and dark regions on the glass. The exhibit states that monochromatic light with a wavelength of \(550 \mathrm{nm}\) is incident on the glass plates and that the plates are sitting in air. The glass has an index of refraction of \(1.51 .\) (a) What is the minimum distance between the two glass plates for one of the dark regions? (b) What is the minimum distance between the two glass plates for one of the light regions? (c) What is the next largest distance between the plates for a dark region? [Hint: Do not worry about the thickness of the glass plates; the thin film is the air between the plates.]
A thin film of oil \((n=1.50)\) of thickness \(0.40 \mu \mathrm{m}\) is spread over a puddle of water \((n=1.33) .\) For which wavelength in the visible spectrum do you expect constructive interference for reflection at normal incidence?
Gratings A grating has exactly 8000 slits uniformly spaced over \(2.54 \mathrm{cm}\) and is illuminated by light from a mercury vapor discharge lamp. What is the expected angle for the third-order maximum of the green line $(\lambda=546 \mathrm{nm}) ?$
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