Chapter 36: Problem 51
A thin slit illuminated by light of frequency \(f\) produces its first dark band at \(\pm\)38.2\(^\circ\) 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 \(\pm\)21.6\(^\circ\). Find the refractive index of the liquid.
Short Answer
Step by step solution
Understanding the Problem
Apply Diffraction Formula
Find Wavelength in the Liquid
Calculate the Refractive Index
Perform the Calculation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Single-Slit Diffraction
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
\[ a \sin \theta = m \lambda \]
where:
- \(a\) is the width of the slit,
- \(\theta\) is the angle of the dark band,
- \(m\) is the order of the minimum (for the first dark band, \(m = 1\)), and
- \(\lambda\) is 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
In an unknown liquid, the wavelength is typically shorter than in air, leading to different diffraction patterns. The relationship between the wavelength in air \(\lambda\) and in a liquid \(\lambda'\) is described by the equation:
\[ \lambda' = \frac{\lambda}{n} \]
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:
\[ n = \frac{\sin \theta_{air}}{\sin \theta_{liquid}} \]
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.