Problem 66 - Physics

Question

Concept Questions 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? Problem 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) 28.8 cm in vacuum. (b) 21.6 cm in water.

Step 1: Trigonometric Relationship

From trigonometry, the distance \(y\) between the central bright fringe and the second-order (or any \(m\)-th order) bright fringe is given by \(y = L \tan \theta\), where \(L\) is the distance from the grating to the screen, and \(\theta\) is the diffraction angle.

Step 2: Relationship from Physics

In diffraction, the angle \(\theta\) for the \(m\)-th order bright fringe is related to the slit separation \(d\), the wavelength \(\lambda\), and the order \(m\) by the equation \(d \sin \theta = m \lambda\).

Step 3: Small Angle Approximation

For small angles, \(\tan \theta \approx \sin \theta\). Thus, \(y = L \tan \theta \approx L \sin \theta\). Using the diffraction condition \(d \sin \theta = m \lambda\), substitute \(\sin \theta\) into the expression for \(y\) to get \(y = \frac{m \lambda L}{d}\).

Step 4: Submerge Apparatus in Water

When submerged in water, the wavelength changes since \(\lambda_{\text{water}} = \frac{\lambda_{\text{vacuum}}}{n_{\text{water}}}\), where \(n_{\text{water}} = 1.33\). Thus, \(y_{\text{water}} = \frac{m \lambda_{\text{water}} L}{d} = \frac{m \lambda L}{n_{\text{water}} d}\). So, the distance \(y\) decreases because the wavelength decreases in water.

Step 5: Calculate \(y\) in Vacuum

Given \(\lambda = 480 \text{ nm} = 480 \times 10^{-9} \text{ m}\), \(d = 5.0 \times 10^{-7} \text{ m}\), \(L = 0.15 \text{ m}\), and \(m = 2\), use the formula \(y = \frac{m \lambda L}{d}\): \[ y = \frac{2 \times 480 \times 10^{-9} \times 0.15}{5.0 \times 10^{-7}} = 0.288 \text{ m} = 28.8 \text{ cm}. \]

Step 6: Calculate \(y\) in Water

Convert the wavelength for water: \(\lambda_{\text{water}} = \frac{480 \times 10^{-9}}{1.33}\). Use the modified formula for \(y\) in water: \[ y_{\text{water}} = \frac{2 \cdot \frac{480 \times 10^{-9}}{1.33} \cdot 0.15}{5.0 \times 10^{-7}} \approx 0.216 \text{ m} = 21.6 \text{ cm}. \]

Key Concepts

Diffraction Angle

The diffraction angle, denoted as \( \theta \), is a fundamental aspect of understanding diffraction patterns. When light passes through a diffraction grating, it bends or diffracts at specific angles where constructive interference occurs. This results in bright and dark fringes on a screen. The position of these fringes is directly related to the diffraction angle. This angle can be determined using the trigonometric relationship for small angles, where \( \tan \theta \approx \sin \theta \). For the situation described, sine can be easily substituted in formulas involving small diffraction angles.

• Smaller angles simplify calculations, making \( \sin \theta \) and \( \tan \theta \) nearly equal.

Wavelength

Wavelength, represented by \( \lambda \), is the distance between consecutive crests of a wave. In a diffraction grating scenario, the wavelength of light determines how the light spreads out upon passing through the grating. Interestingly, when light is submerged in a medium such as water, its velocity decreases due to the medium's refractive index, denoted as \( n \). Therefore, the wavelength in water becomes shorter, calculated as \( \lambda_{\text{water}} = \frac{\lambda_{\text{vacuum}}}{n_{\text{water}}} \).

• A change in medium affects wavelength but not frequency.
• The decrease in wavelength in water causes the diffraction pattern to shrink.

Diffraction Pattern

A diffraction pattern is formed as light passes through a diffraction grating, creating alternating bright and dark fringes on a distant screen. These patterns arise from the constructive and destructive interference of light waves. The bright fringes are where the waves overlap in phase, and their interference is constructive, amplifying the light intensity. The location of these bright fringes is predictable based on the equation \( d \sin \theta = m \lambda \), where \( d \) is the slit spacing, \( m \) is the order number, and \( \lambda \) is the wavelength of light.

• Higher order fringes (larger \( m \)) are spread further apart.
• The central bright fringe (\( m=0 \)) is aligned with the original light beam's path.

Central Bright Fringe

The central bright fringe is the most prominent and central aspect of a diffraction pattern observed on a screen. It is the result of direct light, where light waves from all slits arrive in complete phase alignment, creating a highly bright spot. This fringe is often aligned with the direct path of the incoming light. The distance from this central bright fringe to other fringes can be measured. It is determined using the relationship \( y = \frac{m \lambda L}{d} \). Here, \( L \) is the distance from the grating to the screen, while \( \lambda \) is the wavelength, \( d \) is the separation between the slits, and \( m \) is the fringe order.

• This is the reference point for all other measured bright fringes.
• The position of bright fringes depends on the wavelength and order \( m \).