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A thin slit illuminated by light of frequency f produces its first dark band at ±38.2 in air. When the entire apparatus (slit, screen, and space in between) is immersed in an unknown transparent liquid, the slit's first dark bands occur instead at ±21.6. Find the refractive index of the liquid.

Short Answer

Expert verified
The refractive index of the liquid is approximately 1.672.

Step by step solution

01

Understanding the Problem

We are given that the first dark band is caused by diffraction and it occurs at specific angles in air and in an unknown liquid. We need to find the refractive index of this liquid. The angles given are for the first dark band, which is a characteristic of single-slit diffraction.
02

Apply Diffraction Formula

The formula for the angle of the first dark band in single-slit diffraction is given by asinθ=mλ, where a is the slit width, θ is the angle of the dark band, m is the order of minimum, and λ is the wavelength. For the first minimum, m=1. We use this formula to find relations between the angles in air and in the liquid.
03

Find Wavelength in the Liquid

Given that the angle in air is θair=38.2, so we have asin38.2=λ. In the liquid, θliquid=21.6, so using the same formula asin21.6=λ. The wavelength in the liquid λ relates to the wavelength in air λ by λ=λ/n, where n is the refractive index of the liquid.
04

Calculate the Refractive Index

From both conditions, we know a=λsin38.2=λsin21.6. Substitute λ=λ/n into the second equation, so λnsin21.6=λsin38.2. Cancel λ from both sides and solve for n: n=sin38.2sin21.6.
05

Perform the Calculation

Calculate n using the sine values: n=sin38.2sin21.60.61570.36811.672.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Single-Slit Diffraction
Single-slit diffraction is an important phenomenon in wave optics. It occurs when waves, such as light, pass through a narrow opening and then spread out, creating a pattern of light and dark areas on a screen. The pattern results from constructive and destructive interference of the waves.

In a single-slit setup, the dark bands or minima represent positions where waves cancel out due to destructive interference. This results in less light being observed at these angles. The first minimum is typically observed at a specific angle depending on the wavelength of the light and the width of the slit through which the light passes. The angle at which the first dark band occurs is crucial in determining various properties of the light and the medium it passes through.

One can use this knowledge to determine other factors such as the wavelength of light in different mediums by observing how the angles of the dark bands shift when the setup is moved from one medium to another, as seen in the exercise problem.
Diffraction Formula
The diffraction formula for single-slit diffraction helps us find the angles at which these dark bands appear. The formula is given by:

asinθ=mλ
where:
  • a is the width of the slit,
  • θ is the angle of the dark band,
  • m is the order of the minimum (for the first dark band, m=1), and
  • λ is the wavelength of the light.
This equation allows us to solve for the angle, slit width, or wavelength, depending on the known quantities. In an experimental setup, if the angle and slit width are known, it becomes possible to determine the wavelength of the light.

In different mediums, such as air and a liquid, the light's wavelength changes, which results in a shift in the angles where these minima occur. This principle is employed in the exercise, where the angle shift is used to find the refractive index of a liquid.
Wavelength in Liquid
The wavelength of light changes when it passes from one medium to another. This is due to the refractive index of the medium, which is a measure of how much the speed of light is reduced inside the medium compared to vacuum.

In an unknown liquid, the wavelength is typically shorter than in air, leading to different diffraction patterns. The relationship between the wavelength in air λ and in a liquid λ is described by the equation:

λ=λn
where n is the refractive index of the liquid.

By measuring the angles of diffraction in both air and the liquid, and using the known refractive index of air (which is approximately 1), one can solve for the refractive index of the liquid using:

n=sinθairsinθliquid
This calculation provides a practical way to determine the properties of an unknown transparent medium by using its effect on the behavior of light passing through it.

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Most popular questions from this chapter

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 cm/s on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 m away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at ±61.3 cm from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. (a) How wide is the hole in the barrier? (b) At what other angles do you find no waves hitting the shore?

Why is visible light, which has much longer wavelengths than x rays do, used for Bragg reflection experiments on colloidal crystals? (a) The microspheres are suspended in a liquid, and it is more difficult for x rays to penetrate liquid than it is for visible light. (b) The irregular spacing of the microspheres allows the longerwavelength visible light to produce more destructive interference than can x rays. (c) The microspheres are much larger than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be much longer than the size of the individual scatterers. (d) The microspheres are spaced more widely than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be comparable to the spacing between scattering planes.

An interference pattern is produced by light of wavelength 580 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 mm. (a) If the slits are very narrow, what would be the angular positions of the first-order and second-order, two-slit interference maxima? (b) Let the slits have width 0.320 mm. In terms of the intensity I0 at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 mm. Calculate the wavelength of the light.

If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 nm) at 65.0 from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 nm)?

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