Problem 57 - University Physics with Modern Physics

Question

Two thin parallel slits are made in an opaque sheet of film. When a monochromatic beam of light is shone through them at normal incidence, the first bright fringes in the transmitted light occur in air at \(\pm 18.0^{\circ}\) with the original direction of the light beam on a distant screen when the apparatus is in air. When the apparatus is immersed in a liquid, the same bright fringes now occur at \(\pm 12.6^{\circ} .\) Find the index of refraction of the liquid.

Step-by-Step Solution

Answer

The index of refraction of the liquid is approximately 1.417.

Step 1: Understand the Interference Pattern

The interference pattern created by the double-slit experiment is characterized by alternating bright and dark fringes. Bright fringes occur at angles where the path difference between light from the two slits is a multiple of the wavelength. The angle \( \theta \) to a bright fringe is given by: \[ d \sin \theta = m \lambda \] where \( d \) is the distance between the slits, \( \theta \) is the angle of the fringe, \( m \) is the order of the fringe (\( m = \pm 1, \pm 2, ...\)), and \( \lambda \) is the wavelength of light in the medium.

Step 2: Apply to Air Medium

In air, the first bright fringe (\( m = 1 \)) occurs at \( \theta = 18.0^\circ \). Therefore: \[ d \sin 18.0^\circ = \lambda \] This equation represents the condition for the first bright fringe in air.

Step 3: Apply to Liquid Medium

When the apparatus is immersed in a liquid, the same order of the bright fringe appears at \( \theta = 12.6^\circ \). Therefore, the equation becomes: \[ d \sin 12.6^\circ = \lambda' \] where \( \lambda' \) is the wavelength of light in the liquid. Since the order of the fringe is the same, we equate this to \( m \lambda' = m \lambda_{air} / n \), where \( n \) is the index of refraction of the liquid.

Step 4: Relate Wavelengths in Air and Liquid

The wavelength of light changes when it enters a medium with a refractive index different from 1 (air is approximately 1). The relationship is:\[ \lambda' = \frac{\lambda}{n} \] where \( \lambda \) is the wavelength in air, and \( \lambda' \) is the wavelength in the liquid. From the angles given, \[ \lambda' = \frac{d \sin 12.6^\circ}{1} \] and \[ \lambda = \frac{d \sin 18.0^\circ}{1} \] Therefore, \[ \frac{\lambda}{\lambda'} = \frac{\sin 18.0^\circ}{\sin 12.6^\circ} = n \].

Step 5: Calculate the Index of Refraction

Calculate the index of refraction using the sine values:\[ n = \frac{\sin 18.0^\circ}{\sin 12.6^\circ} \] Using a calculator to find the sine values, we compute:\[ n = \frac{0.3090}{0.2181} \approx 1.417 \].

Step 6: Conclusion

The index of refraction of the liquid is approximately 1.417, indicating the liquid's refractive property compared to air.

Key Concepts

Index of Refraction

The index of refraction refers to how much the speed of light is reduced while traveling through a medium compared to its speed in a vacuum. It is a dimensionless number that provides valuable insight into the optical properties of various materials. For example, air has an index of refraction close to 1, meaning light travels at nearly the same speed as in a vacuum.

In the context of the double-slit experiment, the index of refraction helps determine how the path of light is altered when it enters a non-air medium, like a liquid in this problem. It is calculated using the formula: \[ n = \frac{c}{v} \]where \( n \) is the index of refraction, \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium.

A higher index of refraction indicates that light travels more slowly through the medium, which affects phenomena such as the interference pattern seen in the double-slit experiment.

Interference Pattern

An interference pattern arises when waves, such as light waves, overlap and interact with each other. This pattern of light and dark bands is produced in places where waves interfere constructively or destructively.

In a double-slit experiment, an interference pattern forms when light passes through two thin slits, creating alternating bright and dark spots on a screen. Bright fringes appear where the waves from both slits constructively interfere. This happens when the path difference between the waves is an integer multiple of the wavelength.

• Constructive interference results in bright fringes.
• Destructive interference leads to dark fringes.

The formula used to find the angle for bright fringes in an interference pattern is:\[ d \sin \theta = m \lambda \]where \( d \) is the slit separation, \( \theta \) is the fringe angle, \( m \) is the fringe order, and \( \lambda \) is the wavelength of the light in the medium. Understanding this formula helps us predict where bright fringes will appear based on the medium and light used.

Wavelength in Medium

The wavelength of light changes when it passes from one medium to another, due to the change in speed. In simpler terms, the color or wavelength you see in air might differ in water or glass because the light slows down and its wavelength shortens.

The relationship between the wavelength in air \( \lambda \) and the wavelength in a medium \( \lambda' \) is given by:\[ \lambda' = \frac{\lambda}{n} \]where \( n \) is the index of refraction for the medium.

• The wavelength in the medium \( \lambda' \) is shorter if the index of refraction \( n \) is greater than 1.
• This change affects how bright fringes are positioned in the double-slit setup.

By understanding how the wavelength in a medium varies, students can accurately predict the resulting interference pattern changes when light transitions between different materials.

Bright Fringes

Bright fringes, seen in interference patterns like the double-slit experiment, are bands where light waves add up constructively. These bright spots indicate points where the waves from two slits are perfectly in phase.

To find bright fringes, we examine the phase difference between waves traveling through the slits to the screen. A path difference equivalent to an integer multiple of the wavelength leads to constructive interference, creating the bright bands.

• Path difference: \( m \lambda \), where \( m = 0, \pm 1, \pm 2, \ldots \).
• Formula: \[ d \sin \theta = m \lambda \]
• \( \theta \) is the angle of the bright fringe, \( d \) is the slit distance.

In the provided exercise, bright fringes shift when the apparatus is submerged in a liquid, due to changes in the wavelength of the light. The task is to find how this movement relates to the index of refraction, demonstrating the connection between fringe position and optical properties of the liquid.