Radio waves from a star, of wavelength 250 m, reach a radio telescope by two separate paths, as

step 1 Hello and welcome to this video solution of numerate. Here it's given that radio waves from a st

Radio waves from a star, of wavelength 250 m, reach a radio...

Key Concepts

Path Difference and Geometric Analysis

This key concept involves using geometric relationships and trigonometric analysis to determine the difference in travel paths of waves arriving at a receiver. By relating the angular position of the source, the heights, and distances in the setup, one can derive conditions for interference minima or maxima. This geometric reasoning is essential for solving problems involving multiple wave paths due to reflection.

Destructive Interference

Destructive interference occurs when the phase difference between two interfering waves leads to the cancellation of their amplitudes, resulting in a minimum in the observed intensity. This concept is used to set conditions that relate the physical path differences, the wavelength of the waves, and the geometry of the setup, enabling calculations such as the height of a cliff in reflection problems.

Wave Reflection

Wave reflection occurs when a wave encounters a surface and bounces back rather than being absorbed or transmitted. In contexts like radio astronomy or acoustics, reflection can lead to multiple signal paths, and it is essential to take into account the reflection properties of surfaces, such as whether there is any phase shift upon reflection.

Electromagnetic Wave Propagation

This concept involves the transmission of energy in the form of electromagnetic waves through space. Radio waves, like all electromagnetic radiation, travel in straight lines and can be affected by obstacles or reflective surfaces. Understanding this propagation is fundamental when analyzing how waves from astronomical sources reach an observer via multiple paths.

Interference of Waves

Interference describes the phenomenon where two or more waves superpose to form a resultant wave of greater, lower, or the same amplitude. When waves following different paths meet at a receiver, they can interfere constructively or destructively. This concept is crucial for understanding the observed intensity variations, where phase differences due to path differences determine the interference pattern.

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:04 Here it's given that radio waves from a star of wavelength 250 meters reaches a radio station by two separate paths as shown in this figure, right? so lambda is 250 meters it's given and the two paths are kind of like this here we have the tower or let's say just sort of like this.

00:23 Sorry let me just drop it straight right here and there are two paths using which signals receive sorry here like that and the same after reflecting from the ground right kind of like that, right and uh, right the second is by the reflection of the water right here we have the water right? uh by the ocean right the first minimum of destructive interference occurs when the star is 25 degrees above the horizontal right.

01:09 Uh, now you have to find the height of the cliff, right? so let's say this is the height of the cliff h right.

01:20 Now what you can do is you have got this angle, right? this is theta theta is 25 degrees, right? this is also theta and if you just take a horizontal section, this is also theta, right? now what you need to do is you need to calculate the path difference, right? so if you see here both rays are approaching one is approaching the water another this tower itself right cliff and if you draw this segment then here this section is the only the path difference which is let's say b, right? and let's say this distance is c right and let's also consider this angle phi.

02:05 The phase angle is this phi right now what you have is? delta r or delta r is the path difference for destructive interference, right? you have got m plus one by two times of lambda, right? and this delta r is nothing but c minus b, right? this is a path difference, right? because after reflection the light will travel through the distance c and uh without reflection it will travel through b, right? so c minus b that is equal to m plus one by two lambda now what we need is we need to have c minus this b can be uh changed in terms of c which is c sine theta, right if you check this triangle, right? right.

03:15 Now what you can do is you can consider m to be equal to zero for first uh destructive interference c common then one minus sine theta that is equal to lambda by two, right? next you have to write c in terms of this height, right? so if you consider this triangle then this value of c this is also c, right? this can be written as or maybe we can have this h equal to c sine theta, right? or you can have this c will be equal to h by sine theta times one minus sine theta that is equal to lambda by two fine...

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