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Chapter 25: Problem 32

Ramon has a coherent light source with wavelength \(547 \mathrm{nm} .\) He wishes to send light through a double slit with slit separation of $1.50 \mathrm{mm}\( to a screen \)90.0 \mathrm{cm}$ away. What is the minimum width of the screen if Ramon wants to display five interference maxima?

Short Answer

Expert verified
Answer: The minimum width of the screen needed to display five interference maxima is approximately 32.8 mm.

Step by step solution

01

Determine the formula for the position of bright fringes

We can use the formula for double-slit interference to find the position of bright fringes (interference maxima): $$y = \frac{m \lambda L}{d}$$ where \(y\) is the distance from the central maximum (m=0) to the m-th maximum, \(\lambda\) is the wavelength of light, \(L\) is the distance between the double slits and the screen, \(d\) is the slit separation, and \(m\) is the order number of the maximum (integer value).
02

Calculate the position of the 5th bright fringe

We are asked to find the width of the screen needed to display five interference maxima. The fifth maxima would have an order number \(m = 5\). Plugging in the given values, we get: $$y_5 = \frac{5 (547 \times 10^{-9} \mathrm{m})(0.9 \mathrm{m})}{1.5 \times 10^{-3} \mathrm{m}}$$ Calculating the value, $$y_5 \approx 0.0164 \mathrm{m}$$
03

Calculate the total width of the screen

Since our calculated value, \(0.0164 \mathrm{m}\), represents the distance of the fifth maximum (\(m = 5\)) from the central maximum (\(m = 0\)), the width of the screen would be approximately the sum of the distances of the fifth maximum to the left of the central maximum and the fifth maximum to the right of the central maximum. Therefore, the total width w of the screen would be: $$w = 2 \times y_5$$ Calculating the value, $$w \approx 2 \times 0.0164 \mathrm{m} \approx 0.0328 \mathrm{m}$$ So, the minimum width of the screen needed to display five interference maxima is approximately \(0.0328 \mathrm{m}\) or \(32.8\,\mathrm{mm}\).

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Most popular questions from this chapter

Red light of 650 nm can be seen in three orders in a particular grating. About how many slits per centimeter does this grating have?
Sonya is designing a diffraction experiment for her students. She has a laser that emits light of wavelength \(627 \mathrm{nm}\) and a grating with a distance of \(2.40 \times 10^{-3} \mathrm{mm}\) between slits. She hopes to shine the light through the grating and display a total of nine interference maxima on a screen. She finds that no matter how she arranges her setup, she can see only seven maxima. Assuming that the intensity of the light is not the problem, why can't Sonya display the \(m=4\) interference maxima on either side?
A grating is made of exactly 8000 slits; the slit spacing is $1.50 \mu \mathrm{m} .\( Light of wavelength \)0.600 \mu \mathrm{m}$ is incident normally on the grating. (a) How many maxima are seen in the pattern on the screen? (b) Sketch the pattern that would appear on a screen \(3.0 \mathrm{m}\) from the grating. Label distances from the central maximum to the other maxima.
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 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?
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