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Chapter 34: Problem 62

Coherent monochromatic light passes through parallel slits and then onto a screen that is at a distance \(L=2.40 \mathrm{~m}\) from the slits. The narrow slits are a distance \(d=2.00 \cdot 10^{-5} \mathrm{~m}\) apart. If the minimum spacing between bright spots is \(y=6.00 \mathrm{~cm},\) find the wavelength of the light.

Short Answer

Expert verified
Answer: The wavelength of the light is 10 nm.

Step by step solution

01

Finding the angle θ using the minimum spacing between bright spots, y

We'll use the small angle approximation formula: $$ \theta \approx tan(\theta) = \frac{y}{L} $$ where \(L\) is the distance from the slits to the screen, and \(y\) is the minimum spacing between bright spots. Plugging the given values: $$ \theta = \frac{6.00 \times 10^{-2} \mathrm{~m}}{2.40 \mathrm{~m}} $$ Calculate the value of \(\theta\): $$ \theta = 0.025 $$
02

Finding the order m for the first minimum using the angle θ

For the first minimum (dark spot) in the interference pattern, the path difference between two rays is half a wavelength. Therefore, the corresponding angle will be: $$ \sin(\theta') = \frac{\lambda}{2d} $$ Using the small angle approximation and knowing that \(\theta' = \theta\), $$ \theta' \approx \frac{\lambda}{2d} $$ We can now find the order of the bright spot \(m\) by multiplying both sides of the equation by \(2\): $$ m = 2\theta' = \frac{\lambda}{d} $$ By using the value of \(\theta\) calculated in Step 1, $$ m = 2 \times 0.025 = 0.05 $$
03

Finding the wavelength λ using the constructive interference formula

Now, we can use the constructive interference formula to find the wavelength of the light: $$ \sin(\theta) = \frac{m\lambda}{d} $$ Plugging the calculated values for \(m\), \(d\), and \(\theta\): $$ 0.025 = \frac{0.05\lambda}{2.00 \times 10^{-5} \mathrm{~m}} $$ Solve for \(\lambda\): $$ \lambda = \frac{0.025 \times 2.00 \times 10^{-5} \mathrm{~m}}{0.05} $$ Calculate the value of \(\lambda\): $$ \lambda = 1.00 \times 10^{-5} \mathrm{~m} $$ So, the wavelength of the light is \(1.00 \times 10^{-5} \mathrm{~m}\) or \(10\,\mathrm{nm}\).

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Most popular questions from this chapter

Light of wavelength \(653 \mathrm{nm}\) illuminates a slit. If the angle between the first dark fringes on either side of the central maximum is \(32.0^{\circ},\) What is the width of the slit?

White light is shone on a very thin layer of mica \((n=1.57),\) and above the mica layer, interference maxima for two wavelengths (and no other in between) are seen: one blue wavelength of \(480 \mathrm{nm},\) and one yellow wavelength of \(560 \mathrm{nm} .\) What is the thickness of the mica layer?

Suppose the distance between the slits in a double-slit experiment is \(2.00 \cdot 10^{-5} \mathrm{~m} .\) A beam of light with a wavelength of \(750 \mathrm{nm}\) is shone on the slits. What is the angular separation between the central maximum and the adjacent maximum? a) \(5.00 \cdot 10^{-2} \mathrm{rad}\) b) \(4.50 \cdot 10^{-2} \mathrm{rad}\) c) \(3.75 \cdot 10^{-2} \mathrm{rad}\) d) \(2.50 \cdot 10^{-2} \mathrm{rad}\)

34.35 Monochromatic blue light \((\lambda=449 \mathrm{nm})\) is beamed into a Michelson interferometer. How many fringes move by the screen when the movable mirror is moved a distance \(d=\) \(0.381 \mathrm{~mm} ?\)

A Young's interference experiment is performed with monochromatic green light \((\lambda=540 \mathrm{nm}) .\) The separation between the slits is \(0.100 \mathrm{~mm},\) and the interference pattern on a screen shows the first side maximum \(5.40 \mathrm{~mm}\) from the center of the pattern. How far away from the slits is the screen?
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