In a double-slit experiment, He-Ne laser light of wavelength \(633 \mathrm{nm}\) produced an interference pattern on a screen placed at some distance from the slits. When one of the slits was covered with a thin glass slide of thickness \(12.0 \mu \mathrm{m},\) the central fringe shifted to the point occupied earlier by the 10 th dark fringe (see figure). What is the refractive index of the glass slide? (a) Without the glass slide (b) With glass slide

Short Answer

Expert verified
Answer: The refractive index of the glass slide is approximately 1.53.

Step by step solution

01

Identifying known values and the formula for path difference

We are given: - Wavelength of light (λ) = 633 nm = 633 x 10^-9 m - Thickness of the glass slide (t) = 12.0 μm = 12.0 x 10^-6 m - The central fringe shifts to the previous 10th dark fringe when the glass slide is placed. We need to find the path difference (Δ) caused by the glass slide. The formula for path difference in a double-slit experiment is given by: Δ = mλ where m = the order of the dark fringe (in this case, m=10).
02

Calculate the path difference

Using the given values and the path difference formula, we can calculate the path difference: Δ = 10λ = 10 * (633 * 10^-9 m) = 6.33 * 10^-6 m
03

Determine the extra distance traveled in the glass slide

When the glass slide is placed over one slit, the light from that slit has to travel an extra distance (d) through the slide compared to the other slit. This extra distance is given by: d = (n - 1) t where n = refractive index of the glass slide.
04

Use the path difference to find the refractive index

We have the path difference (Δ) and the extra distance (d) formulas: Δ = 6.33 * 10^-6 m d = (n - 1) t Since the path difference is equal to the extra distance traveled in the glass slide, we can equate the two formulas: 6.33 * 10^-6 m = (n - 1) * (12.0 * 10^-6 m) Now, we can solve for the refractive index (n): n = (6.33 * 10^-6 m / (12.0 * 10^-6 m)) + 1
05

Calculate the refractive index

Plugging the values into the equation, we get: n = (6.33 / 12) + 1 = 1.5275 Thus, the refractive index of the glass slide is approximately 1.53.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Can light pass through a single slit narrower than its wavelength? If not, why not? If so, describe the distribution of the light beyond the slit.

The thermal stability of a Michelson interferometer can be improved by submerging it in water. Consider an interferometer that is submerged in water, measuring light from a monochromatic source that is in air. If the movable mirror moves a distance of \(d=0.200 \mathrm{~mm},\) exactly \(N=800\) fringes move by the screen. What is the original wavelength (in air) of the monochromatic light?

The Hubble Space Telescope (Figure 34.33 ) is capable of resolving optical images to an angular resolution of \(2.80 \cdot 10^{-7}\) rad with its 2.40 -m mirror. How large would a radio telescope have to be in order to image an object in the radio spectrum with the same resolution, assuming the wavelength of the waves is \(10.0 \mathrm{~cm} ?\)

At the Long-baseline Interferometer Gravitationalwave Observatory (LIGO) facilities in Hanford, Washington, and Livingston, Louisiana, laser beams of wavelength \(550.0 \mathrm{nm}\) travel along perpendicular paths \(4.000 \mathrm{~km}\) long. Each beam is reflected along its path and back 100 times before the beams are combined and compared. If a gravitational wave increases the length of one path and decreases the other, each by 1.000 part in \(10^{21}\), what is the phase difference between the two beams as a result?

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) \(2.50 \cdot 10^{-2} \mathrm{rad}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free