A radio telescope consists of two antennae separated by a distance of 235 m. Both antennae are tuned to 26 MHz. The signals from the two antennae are fedinto a common amplifier, but one signal is passed through a phase adjuster that delays the phase by a chosen amount, 6, so that the telescope effectively"looks" in different directions. When 8=0, plane radio waves that are incident vertically on the antennae produce signals that add constructively at theamplifier. What is the minimum phase delay that should be introduced in order that signals coming from an angle 0 = 18 degrees from the vertical will addconstructively at the amplifier? Give your answer as a positive number in radians, and use 3 x 108 m/s for the speed of light.NumberBackQuestion Menu -Submit AssignmentQuit & Save

Here we have to use some basic formula here

Answered: A radio telescope consists of two…

Similar questions

In the figure, two radio-frequency point sources S1 and S2, separated by distance d = 1.6 m, are radiating in phase with A = 0.4 m. A detector moves in a large circular path around the two sources in a plane containing them. How many maxima does it detect? Number x S₁ S2 Units
In a Young's double-slit experiment, a set of parallel slits with a separation of 0.108 mm is illuminated by light having a wavelength of 584 nm and the interference pattern observed on a screen 3.50 m from the slits. (a) What is the difference in path lengths from the two slits to the location of a third order bright fringe on the screen? answer in ?m (b) What is the difference in path lengths from the two slits to the location of the third dark fringe on the screen, away from the center of the pattern? answer in ?m
Two slits separated by a distance of d=0.150 mm are located at a distance of D=810. mm from a screen. The screen is oriented parallel to the plane of the slits. The slits are illuminated by a coherent light source with a wavelength of λ=549×10−6 mm. The interference pattern shows a peak at the center of the screen (m=0) and then alternating minima and maxima. (a) What is the path length difference in millimeters between the two waves from the two slits at the first (m=1) maximum on the screen? (b) What is the path length difference in millimeters between the two waves from the two slits at the first (m=0) minimum on the screen? (c) Calculate the distance on the screen between the central maximum (m=0) and the first (m=1) maximum. You can assume sinθ≈tanθ≈θ, with θ expressed in radians. Give your answer in millimeters.
An interference filter has a dielectric layer (refractive index 1.34) with a thickness of 0.805 μm. If the determination is to be based on the first-order interference, calculate the wavelength (nm) can be transmitted?
The full width at half-maximum (FWHM) of a central diffraction maximum is defined as the angle between the two points in the pattern where the intensity is one-half that at the center of the pattern. (See figure (b).) (a) Does the intensity drop to one-half the maximum value when sin²α = a²/2? (b) Is a = 1.39 rad (about 80°) a solution to the transcendental equation of (a)? (c) Is the FWHM AÐ = 2sin¹(0.442 A/a), where a is the slit width? Calculate the FWHM of the central maximum for slit width (d) 1.17 A, (e) 5.03 A, and (f) 11.7 A. 20 20 Relative intensity 15 10 0.8 0.6 a=2 0.4 0.2 5 05 8 (degrees) (a) 10 15 20 20 Relative intensity 1.0 0.8 0.6 -A0- 0.4 0.2 a= 52 20 15 10 5 0 5 10 15 20 (degrees) (b)
In a Young's double-slit experiment, the angle that locates the second dark fringe on either side of the central bright fringe is 4.0 degrees. Find the ratio d/A of the slit separation d to the wavelength of the light. Number Units
In a single-slit diffraction pattern, the central fringe is 480 times as wide as the slit. The screen is 16,000 times farther from the slit than the slit is wide. What is the ratio X/W, where is the wavelength of the light shining through the slit and W is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that sin is approximately equal to tan (). Number Units
A spacer is cut from a playing card of thickness 2.88 x 10-4 m and used to separate one end of two rectangular, optically flat, 3.10 cm long glass plates with n = 1.65, as in the figure below. Laser light at 633 nm shines straight down on the top plate. HINT (a) Count the number of phase reversals for the interfering waves. 909.9 (b) Calculate the separation (in m) between dark interference bands observed on the top plate. Need Help? Read It