Microwaves with λ = 1 mm are incident upon a 10 cm slit. At a distance of 1 m from the slit, what is the width of the slit's image?
The answer is supposed to be 10 cm.
I can't seem to get that though. Can someone explain?
The pattern beyond the slit is usually not called an image; it is called a diffraction pattern. At this close range, you may be dealing with near-field Fresnel rather than conventional Fraunhofer scattering. You need to review the differences and apply the correct formulas, which are quite complex for Fresnel scattering.
(λ/D) is rather small in this case (0.01), so I expect there to be a bright spot about 10 cm wide 1 m beyond the slit, with some fringes at the edges.
I just don't have enough time to review all this for you. Sorry
Apparently your book answer agrees with mine, but I should have used the terms Fresnel and Franhofer DIFFRACTION, not scattering. The type of diffraction that you have depends upon the dimensionless ratio
D/sqrt(λ R) If the ratio is <1, you have Fraunhofer scattering with uniformly spaced light and dark bands. In your case that ratio is much larger than 1. Most radiation will follow geometrical-optics ray paths to the near field screen
To determine the width of the slit's image, we can use the concept of diffraction. Diffraction occurs when a wave encounters an obstacle or slit that is comparable in size to the wavelength of the wave. In this case, microwaves with a wavelength of λ = 1 mm encounter a 10 cm slit.
Diffraction causes the wave to spread out and create an interference pattern on a screen placed at a distance. The angular width of the central maximum in the diffraction pattern can be approximated using the formula:
θ ≈ λ / a
Where:
θ is the angular width of the central maximum
λ is the wavelength of the wave
a is the width of the slit
Using the given values, we have:
λ = 1 mm = 0.001 m
a = 10 cm = 0.1 m
Plugging these values into the formula, we get:
θ ≈ 0.001 / 0.1 = 0.01 radians
To find the width of the slit's image, we need to calculate the linear width of the central maximum on the screen. We can use the tangent function to relate the angular width and the linear width:
tan(θ) = (width of the image) / (distance from the slit to the screen)
Rearranging the formula to solve for the width of the image, we have:
(width of the image) = tan(θ) * (distance from the slit to the screen)
In this case, the distance from the slit to the screen is 1 m. Plugging in the values, we get:
(width of the image) = tan(0.01) * 1 = 0.0175 m
Converting this to centimeters, we get:
(width of the image) = 0.0175 m * 100 cm/m = 1.75 cm
Therefore, the width of the slit's image is approximately 1.75 cm, not 10 cm as stated in the answer you provided. Double-check the given information and calculations to ensure accuracy.