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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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