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Ch 36: Diffraction
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
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Capitolo 36, Problema 15b

A slit 0.240 mm wide is illuminated by parallel light rays of wavelength 540 nm. The diffraction pattern is observed on a screen that is 3.00 m from the slit. The intensity at the center of the central maximum (θ = 0°) is 6.00 x 10-6 W/m2. What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

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1
Understand the problem: This is a single-slit diffraction problem. The intensity of light at a point on the screen is determined by the diffraction pattern, which depends on the slit width, wavelength, and the angle of observation. The goal is to find the intensity at a point midway between the central maximum and the first minimum.
Recall the formula for the intensity in a single-slit diffraction pattern: \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \), where \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \), \( a \) is the slit width, \( \lambda \) is the wavelength, and \( \theta \) is the angle of observation. \( I_0 \) is the intensity at the central maximum.
Determine the angle \( \theta \) midway between the central maximum and the first minimum. The first minimum occurs when \( \sin(\theta) = \frac{\lambda}{a} \). Midway between the central maximum and the first minimum corresponds to \( \sin(\theta) = \frac{1}{2} \cdot \frac{\lambda}{a} \). Substitute \( \lambda = 540 \; \text{nm} = 540 \times 10^{-9} \; \text{m} \) and \( a = 0.240 \; \text{mm} = 0.240 \times 10^{-3} \; \text{m} \) to calculate \( \sin(\theta) \).
Calculate \( \beta \) for this angle using \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \). Substitute the values of \( a \), \( \lambda \), and \( \sin(\theta) \) into the equation to find \( \beta \).
Substitute \( \beta \) into the intensity formula \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \). Use \( I_0 = 6.00 \times 10^{-6} \; \text{W/m}^2 \) to calculate the intensity at the specified point. This will give the desired result.

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Diffraction

Diffraction is the bending of waves around obstacles and the spreading of waves when they pass through narrow openings. In the context of light, diffraction patterns arise when light encounters a slit, leading to the formation of bright and dark regions on a screen due to constructive and destructive interference of the light waves.
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Diffraction

Intensity of Light

The intensity of light is defined as the power per unit area carried by a wave. It is proportional to the square of the amplitude of the wave and is measured in watts per square meter (W/m²). In diffraction patterns, intensity varies with position, being highest at the center and decreasing towards the minima.
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Wave Intensity

Position of Minima in Single-Slit Diffraction

In single-slit diffraction, the positions of the minima can be calculated using the formula a sin(θ) = mλ, where 'a' is the slit width, 'θ' is the angle of the minima, 'm' is the order of the minimum (m = ±1, ±2,...), and 'λ' is the wavelength of light. The first minimum occurs at the angle where the first dark fringe appears, which is crucial for determining intensity at specific points on the screen.
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Single Slit Diffraciton
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