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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

In a double-slit interference experiment, the wavelength is \(475 \mathrm{nm}\), the slit separation is \(0.120 \mathrm{mm},\) and the screen is $36.8 \mathrm{cm}$ away from the slits. What is the linear distance between adjacent maxima on the screen? [Hint: Assume the small-angle approximation is justified and then check the validity of your assumption once you know the value of the separation between adjacent maxima.] (tutorial: double slit 1 ).
A Michelson interferometer is set up using white light. The arms are adjusted so that a bright white spot appears on the screen (constructive interference for all wavelengths). A slab of glass \((n=1.46)\) is inserted into one of the arms. To return to the white spot, the mirror in the other arm is moved $6.73 \mathrm{cm} .$ (a) Is the mirror moved in or out? Explain. (b) What is the thickness of the slab of glass?
White light containing wavelengths from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\) is shone through a grating. Assuming that at least part of the third-order spectrum is present, show that the second-and third-order spectra always overlap, regardless of the slit separation of the grating.
A thin layer of an oil \((n=1.60)\) floats on top of water \((n=1.33) .\) One portion of this film appears green \((\lambda=510 \mathrm{nm})\) in reflected light. How thick is this portion of the film? Give the three smallest possibilities.
The radio telescope at Arecibo, Puerto Rico, has a reflecting spherical bowl of \(305 \mathrm{m}(1000 \mathrm{ft})\) diameter. Radio signals can be received and emitted at various frequencies with appropriate antennae at the focal point of the reflecting bowl. At a frequency of \(300 \mathrm{MHz}\), what is the angle between two stars that can barely be resolved? (Tutorial:radio telescope).
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