Problem 66 - Physics

Question

Light shines on a diffraction grating, and a diffraction pattern is produced on a viewing screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) From trigonometry, how is the distance \(y\) from the central bright fringe to the second-order bright fringe related to the diffraction angle \(\theta\) and the distance \(L\) between the grating and the screen? (b) From physics, how is \(\theta\) related to the order \(m\) of the bright fringe, the wavelength \(\lambda\) of the light, and the separation \(d\) between the slits? (c) In this problem, the angle \(\theta\) is small (less than a few degrees). When the angle is small, \(\tan \theta\) is approximately equal to \(\sin \theta\), or \(\tan \theta \approx \sin \theta\). Using this approximation, obtain an expression for \(y\) in terms of \(L, m, \lambda\), and \(d\). (d) If the entire apparatus in the drawing is submerged in water, would you expect the distance \(y\) to increase, decrease, or remain unchanged? Why? Light of wavelength \(480 \mathrm{~nm}\) (in vacuum) is incident on a diffraction grating that has a slit separation of \(5.0 \times 10^{-7} \mathrm{~m}\). The distance between the grating and the viewing screen is \(0.15 \mathrm{~m}\). (a) Determine the distance \(y\) from the central bright fringe to the second-order bright fringe. (b) If the entire apparatus is submerged in water \(\left(n_{\text {water }}=1.33\right)\), what is the distance \(y ?\) Be sure your answer is consistent with part (d) of the Concept Questions.

Step-by-Step Solution

Answer

(a) 0.28 m; (b) 0.21 m, distance decreases when submerged in water.

Step 1: Relate y to θ and L

From trigonometry, the distance from the central fringe to the second-order bright fringe is given by:\[ y = L \tan(\theta) \]This relates the distance \(y\) on the screen to the diffraction angle \(\theta\) and the distance \(L\) from the grating to the screen.

Step 2: Relate θ to m, λ, and d

From the physics of diffraction, the angle \(\theta\) for bright fringes is given by:\[ d \sin(\theta) = m\lambda \]where \(m\) is the order of the fringe, \(\lambda\) is the wavelength, and \(d\) is the slit separation.

Step 3: Use Small Angle Approximation

For small angles, \(\tan \theta \approx \sin \theta\). Thus, we can write:\[ y = L \sin(\theta) \]Substituting \(\sin(\theta)\) from the diffraction condition, we get:\[ y = L \frac{m\lambda}{d} \]

Step 4: Find y for Second Order

Given, \(\lambda = 480 \times 10^{-9} \text{ m} \), \(d = 5.0 \times 10^{-7} \text{ m} \), and \(L = 0.15 \text{ m} \). For the second-order fringe \(m = 2\),\[ y = 0.15 \times \frac{2 \times 480 \times 10^{-9}}{5.0 \times 10^{-7}} = 0.28 \text{ m} \]

Step 5: Effect of Water on y

Submerging the apparatus in water changes the wavelength. The wavelength in water is given by:\[ \lambda_{\text{water}} = \frac{\lambda}{n} = \frac{480 \times 10^{-9}}{1.33} \approx 361.2 \times 10^{-9} \text{ m} \]Thus, the new distance \(y\) is:\[ y_{\text{water}} = L \frac{m\lambda_{\text{water}}}{d} = 0.15 \times \frac{2 \times 361.2 \times 10^{-9}}{5.0 \times 10^{-7}} \approx 0.21 \text{ m} \]Submerging the apparatus decreases the distance \(y\).

Key Concepts

Diffraction Angle

When light is shone on a diffraction grating, it bends around the edges of the slits, creating a pattern on a screen. The angle at which this light deviates from its original path is referred to as the diffraction angle, denoted by \( \theta \). This angle is not arbitrarily formed but arises from the underlying physics: namely the geometry of the setup and the wavelength of the light involved.
In the context of a diffraction grating, the position of bright fringes, which are spots of constructive interference, is fundamentally related to this angle. For a given order of fringe \( m \), which indicates the sequence of the visible bright fringe starting from the center, the diffraction angle is found using the formula:
\[ d \sin(\theta) = m\lambda \]
Here, \( d \) represents the distance between the slits (slit separation), \( \lambda \) is the wavelength of the light, and \( \theta \) is solved for as \( \sin^{-1}\left(\frac{m\lambda}{d}\right) \). This equation explains that the angle increases with both increasing wavelength and order of the bright fringe. Additionally, the smaller the slit separation, the greater the angle, resulting in a wider spread of the diffraction pattern.
Understanding how this angle changes helps in calculating where the light will hit the screen, allowing us to see the intricate and beautiful diffraction patterns so often associated with this phenomenon.

Small Angle Approximation

In optics, small angles offer simplifications that make calculations more manageable. Specifically, the small angle approximation is a technique where the trigonometric functions \( \sin \theta \) and \( \tan \theta \) can be considered approximately equal for small angles (usually less than a few degrees). This can be expressed as:
\[ \tan \theta \approx \sin \theta \approx \theta \]
(in radians), and it is highly useful in the mathematical analysis of diffraction patterns.
The approximation applies well in many practical scenarios involving diffraction gratings. When dealing with the diffraction angle \( \theta \), which is small, simplifying \( \tan \theta \approx \sin \theta \) helps convert mathematical expressions.
For instance, the displacement \( y \) from the central bright fringe can be written as:
\[ y = L \tan(\theta) \]
Using the small angle approximation, this becomes:
\[ y = L \sin(\theta) \]
Substituting \( \sin(\theta) = \frac{m\lambda}{d} \), we obtain:
\[ y = L \frac{m\lambda}{d} \]
This simplification makes it easier to relate physical attributes like the wavelength and slit distance directly to the observed pattern without cumbersome calculations, ensuring easier learning and application.

Wavelength and Medium

The behavior of light depends intricately on its wavelength and the medium it travels through. Wavelength, denoted by \( \lambda \), is the distance between consecutive peaks of a wave, which for light, dictates its color in the visible spectrum. In the context of diffraction, changes in wavelength significantly influence the diffraction pattern produced.
Moreover, the medium plays a pivotal role in altering this wavelength. When light travels through a medium other than a vacuum, its speed and thus its wavelength are affected. This is due to the refractive index \( n \) of the medium, a measure of how much the speed of light is reduced inside the material. The relationship is given by:
\[ \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{n} \]
For example, submerging an optical setup in water (with a typical refractive index \( n \approx 1.33 \)) causes the wavelength to shorten. This directly affects the diffraction pattern, decreasing the distance \( y \) between central and higher-order bright fringes since they are more closely spaced.
Therefore, understanding the connection between wavelength and the medium is pivotal for applications that utilize or analyze diffraction, as it allows accurate predictions and adjustments of the resulting patterns in different environments.