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At rest, a light source emits 532nmlight. (a) As it moves along the line connecting it and Earth. observers on Earth see412nm . What is the source's velocity (magnitude and direction)? (b) Were it to move in the opposite direction at the same speed. what wavelength would be seen? (c) Were it to circle Earth at the same speed. what wavelength would be seen?

Short Answer

Expert verified

(a)The velocity of the source is 0.25cand it is approaching towards the source.

(b)The wavelength of the light when the light is receding is687鈥塶尘 .

(c)The wavelength of the light for circling the Earth549鈥塶尘 .

Step by step solution

01

Write the given data from the question.

The wavelength of source,source=532鈥塶尘

The wavelength of observer,obs=412鈥塶尘

02

 Step 2: Determine the formulas to calculate the source velocity, wavelength in opposite direction and wavelength of light for circling Earth.

The expression to calculate the observer frequency in terms of source frequency, speed and speed of light is given as follows.

fobs=fsource1-(vc)21+vccos

Here,is the angle between the source of light and observer.

The expression between the frequency, wavelength and speed of light is given as follows.

=cf

Here,is the wavelength, cis the speed of light and role="math" localid="1658810978725" fis the frequency.

03

(a) Calculate the source velocity.

Determine the wavelength equation for the observer.

obs=cfobs

Substitutefsource1(vc)21+vccosforfobsinto above equation.

obs=cfsource1(vc)21+vccosobs=cfsource1+vccos1(vc)2

Substitutesourcefor cfsourceinto above equation.

obs=source1+vccos1(vc)2 鈥(颈)

The observed wavelength is shorter than the wavelength in the rest frame. Therefore, light is approaching.The angle between the source and observer when the source moving towards the observe is equal to 180.

Substitute 180forinto equation (i).

obs=source1+vccos1801(vc)2obs=source1vc1(vc)2obs1(vc)2=source(1vc)

Take a square of both the sides of the above equation.

0bs2[1(vc)2]=source2(1vc)20bs2[(1vc)(1+vc)]=source2(1vc)20bs2(1+vc)=source2(1vc)0bs2+vc0bs2=source2source2vc

Solve further as,

vc0bs2+source2vc=source20bs2vc(0bs2+source2)=source20bs2vc=source20bs20bs2+source2

Substitute532鈥塶尘 for sourceand412鈥塶尘 forobs into above equation.

vc=532241225322+4122vc=113280452768vc=0.25v=0.25c

Hence the velocity of the source is 0.25cand it is approaching towards the source.

04

(b) Determine the wavelength of the light for the reverse direction.

The speed of the light would be the same in the opposite direction but the angle between the observer and source is0.

Calculate the wavelength of the light,

Substitute 532鈥塶尘forsource, 0for and0.25cfor vinto equation (i).

source=5321091+0.25cccos01(0.25cc)2source=5321091+0.2510.0625source=5321091.29source=687鈥塶尘

Hence the wavelength of the light when the light is receding is687鈥塶尘 .

05

(c) Determine the wavelength of the light for It is circling the Earth.

The speed of the light would be the same but for circling the Earth the angle between the observer and source is 90.

Calculate the wavelength of the light,

Substitute 532鈥塶尘forsource ,90for and0.25c forv into equation (i).

source=5321091+0.25cccos901(0.25cc)2source=532109110.0625source=5321091.032source=549鈥塶尘

Hence the wavelength of the light for circling the Earth 549鈥塶尘.

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