Multiple Choice
A single slit is illuminated by coherent monochromatic light at . On a screen behind the slit, a central bright spot is observed to have a width of . What is the width of the single slit?
627
views
34. Wave Optics
Single Slit Diffraction
Problem 22
Textbook Question
Light of 630 nm wavelength illuminates a single slit of width 0.15 mm. FIGURE EX33.22 shows the intensity pattern seen on a screen behind the slit. What is the distance to the screen?
Verified step by step guidance1
Step 1: Understand the problem. The question involves diffraction through a single slit, and we are tasked with finding the distance to the screen. The intensity pattern shown in the graph is a result of the diffraction of light of wavelength 630 nm through a slit of width 0.15 mm.
Step 2: Recall the formula for the angular position of minima in single-slit diffraction: \( \sin \theta = \frac{m \lambda}{a} \), where \( m \) is the order of the minima (e.g., \( m = 1 \) for the first minimum), \( \lambda \) is the wavelength of light, and \( a \) is the slit width.
Step 3: Relate the angular position \( \theta \) to the linear position \( x \) on the screen using the small angle approximation: \( \tan \theta \approx \sin \theta \approx \frac{x}{L} \), where \( L \) is the distance to the screen and \( x \) is the distance from the central maximum to the first minimum on the screen.
Step 4: From the graph, observe that the distance \( x \) from the central maximum to the first minimum is approximately 0.5 cm (or 0.005 m). Substitute \( \sin \theta \approx \frac{m \lambda}{a} \) and \( \sin \theta \approx \frac{x}{L} \) into the equation to solve for \( L \): \( L = \frac{x a}{m \lambda} \). Use \( m = 1 \), \( \lambda = 630 \times 10^{-9} \) m, \( a = 0.15 \times 10^{-3} \) m, and \( x = 0.005 \) m.
Step 5: Perform the substitution and simplify the expression for \( L \). This will give the distance to the screen. Ensure units are consistent throughout the calculation.
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Single Slit Diffraction
Single slit diffraction occurs when light passes through a narrow opening, causing the light waves to spread out and create an interference pattern. The width of the slit and the wavelength of the light determine the pattern's characteristics, including the position and intensity of the bright and dark fringes observed on a screen.
Recommended video:
Guided course
12:33
Single Slit Diffraciton
Wavelength and Frequency
Wavelength is the distance between successive peaks of a wave, while frequency is the number of peaks that pass a point in a given time. In this context, the wavelength of 630 nm (nanometers) is crucial for calculating the diffraction pattern, as it influences the angle and spacing of the intensity maxima and minima on the screen.
Recommended video:
Guided course
05:08Circumference, Period, and Frequency in UCM
Intensity Pattern
The intensity pattern observed on the screen is a result of the constructive and destructive interference of light waves emanating from different parts of the slit. The central maximum is the brightest point, with subsequent maxima and minima decreasing in intensity, which can be quantitatively analyzed to determine the distance to the screen based on the slit width and wavelength.
Recommended video:
Guided course
02:32Wave Intensity
12:33mWatch next
Master Single Slit Diffraciton with a bite sized video explanation from Patrick
Start learningRelated Videos
Related Practice