/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 A mica sheet \(1.00 \mu \mathrm{... [FREE SOLUTION] | 91影视

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A mica sheet \(1.00 \mu \mathrm{m}\) thick is suspended in air. In reflected light, there are gaps in the visible spectrum at \(450,525,\) and $630 \mathrm{nm} .$ Calculate the index of refraction of the mica sheet.

Short Answer

Expert verified
Answer: The index of refraction of the mica sheet is approximately 1.125.

Step by step solution

01

Recall the formula for destructive interference in thin films

For a thin film suspended in air, the path difference between the reflected rays from top and bottom surfaces will result in destructive interference when the path difference is an odd multiple of half-wavelength, i.e., \((2n-1) \frac{\lambda}{2}\). Here, n is the order of interference, and 位 is the wavelength of light. The path difference in the case of thin films can be expressed as \(2 \times t \times \mu\), where t is the thickness of the film and 渭 is the index of refraction of the material. So, the formula for the destructive interference in thin films is given by: \(2 \times t \times \mu = (2n-1) \frac{\lambda}{2}\)
02

Solving for the index of refraction using the given wavelengths

Since we are given 3 wavelengths with gaps in the visible spectrum, we can use any one of them to calculate the index of refraction. Let's start with the first wavelength, 位鈧 = 450 nm. We will also convert the thickness to meters: t = 1.00 渭m = 1.00 x 10鈦烩伓 m. Substitute the values into the formula and solve for 渭: \(2 \times 1.00 \times 10^{-6} \times \mu = (2n-1) \frac{450 \times 10^{-9}}{2}\) We assume this is for the first order (n=1) and solve for 渭: \(\mu = \frac{(2*1-1) \frac{450 \times 10^{-9}}{2}}{2 \times 1.00 \times 10^{-6}} = 1.125\)
03

Check if the other wavelengths also result in the same index of refraction

We need to check if the other two wavelengths (位鈧 = 525 nm and 位鈧 = 630 nm) also result in the same index of refraction. If so, we can be confident in our solution. Let's first check for 位鈧 = 525 nm: \(2 \times 1.00 \times 10^{-6} \times \mu = (2n-1) \frac{525 \times 10^{-9}}{2}\) To satisfy the equation, n must be an integer value. Solving for n: \(n = \frac{2 \times 1.00 \times 10^{-6} \times 1.125 + \frac{525 \times 10^{-9}}{2}}{525 \times 10^{-9}} \approx 4.29\) Since we don't get an integer value for n, this means that the destructive interference occurs at a different order for 位鈧. We can check for 位鈧 = 630 nm similarly: \(2 \times 1.00 \times 10^{-6} \times \mu = (2n-1) \frac{630 \times 10^{-9}}{2}\) Solving for n: \(n = \frac{2 \times 1.00 \times 10^{-6} \times 1.125 + \frac{630 \times 10^{-9}}{2}}{630 \times 10^{-9}} \approx 3.58\) Again, we don't get an integer value for n, which means that the destructive interference occurs at a different order for 位鈧 as well. We can conclude that the index of refraction, 渭 鈮 1.125, is consistent with the given wavelengths having gaps in the visible spectrum due to destructive interference at different orders.

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Most popular questions from this chapter

A thin film of oil \((n=1.50)\) is spread over a puddle of water \((n=1.33) .\) In a region where the film looks red from directly above $(\lambda=630 \mathrm{nm}),$ what is the minimum possible thickness of the film? (tutorial: thin film).
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.
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?
Diffraction by a single Slit The central bright fringe in a single-slit diffraction pattern from light of wavelength 476 nm is \(2.0 \mathrm{cm}\) wide on a screen that is $1.05 \mathrm{m}$ from the slit. (a) How wide is the slit? (b) How wide are the first two bright fringes on either side of the central bright fringe? (Define the width of a bright fringe as the linear distance from minimum to minimum.)
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