Question

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

The refractive index of the liquid is approximately 1.672.

Step-by-step solution

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.

Apply Diffraction Formula

The formula for the angle of the first dark band in single-slit diffraction is given by \( a \sin \theta = m \lambda \), where \(a\) is the slit width, \(\theta\) is the angle of the dark band, \(m\) is the order of minimum, and \(\lambda\) 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.

Find Wavelength in the Liquid

Given that the angle in air is \(\theta_{air} = 38.2^\circ\), so we have \( a \sin 38.2^\circ = \lambda \). In the liquid, \(\theta_{liquid} = 21.6^\circ\), so using the same formula \( a \sin 21.6^\circ = \lambda' \). The wavelength in the liquid \(\lambda'\) relates to the wavelength in air \(\lambda\) by \( \lambda' = \lambda/n \), where \(n\) is the refractive index of the liquid.

Calculate the Refractive Index

From both conditions, we know \( a = \frac{\lambda}{\sin 38.2^\circ} = \frac{\lambda'}{\sin 21.6^\circ} \). Substitute \(\lambda' = \lambda/n\) into the second equation, so \( \frac{\lambda}{n \cdot \sin 21.6^\circ} = \frac{\lambda}{\sin 38.2^\circ} \). Cancel \( \lambda \) from both sides and solve for \( n \): \( n = \frac{\sin 38.2^\circ}{\sin 21.6^\circ} \).

Perform the Calculation

Calculate \(n\) using the sine values: \( n = \frac{\sin 38.2^\circ}{\sin 21.6^\circ} \approx \frac{0.6157}{0.3681} \approx 1.672 \).

Key concepts

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:

\[ a \sin \theta = m \lambda \]
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.

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 \(\lambda\) 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.