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Chapter 37: 34P (page 1147)

A sodium light source moves in a horizontal circle at a constant speed of 0.100c while emitting light at the proper wavelength of λo=589nm. Wavelength λis measured for that light by a detector fixed at the center of the circle. What is the wavelength shift λ-λo?

Short Answer

Expert verified

The wavelength shift for sodium light source is 2.97 nm.

Step by step solution

01

Identification of given data

The proper wavelength of emitting light is λo=589nm

The speed of sodium light source is u=0.1c

The wavelength shift is the variation in the wavelength of light due to relative movement of source and observer in different frames. It is calculated by the wavelength shift formula.

02

Determination of wavelength shift for sodium light source

The wavelength shift for sodium light source is given as:

λ-λo=λo11-vc2-1

Substitute all the values in the above equation.

λ-λo=589nm11-0.1cc2-1λ-λo=2.97nm

Therefore, the wavelength shift for sodium light source is 2.97 nm.

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Most popular questions from this chapter

A space traveler takes off from Earth and moves at speed 0.9900ctoward the star Vega, which is 26.00lydistant. How much time will have elapsed by Earth clocks (a) when the traveler reaches Vega and (b) when Earth observers receive word from the traveler that she has arrived? (c) How much older will Earth observers calculate the traveler to be (measured from her frame) when she reaches Vega than she was when she started the trip?

Superluminal jets. Figure 37-29a shows the path taken by a knot in a jet of ionized gas that has been expelled from a galaxy. The knot travels at constant velocity v→ at angleθ from the direction of Earth. The knot occasionally emits a burst of light, which is eventually detected on Earth. Two bursts are indicated in Fig. 37-29a, separated by timet as measured in a stationary frame near the bursts. The bursts are shown in Fig. 37-29b as if they were photographed on the same piece of film, first when light from burst 1 arrived on Earth and then later when light from burst 2 arrived. The apparent distanceD¯app traveled by the knot between the two bursts is the distance across an Earth-observer’s view of the knot’s path. The apparent timeT¯app between the bursts is the difference in the arrival times of the light from them. The apparent speed of the knot is then V¯app=D¯app/T¯app. In terms of v, t, andθ , what are (a)D¯app and (b)T¯app ? (c) EvaluateV¯app forv=0.980c and θ=30.0∘. When superluminal (faster than light) jets were first observed, they seemed to defy special relativity—at least until the correct geometry (Fig. 37-29a) was understood.

Reference frameS'is to pass reference frame Sat speed valong the common direction of the x'and xaxes, as in Fig. 37-9. An observer who rides along with frame s'is to count off 25son his wristwatch. The corresponding time interval ∆tis to be measured by an observer in frame s. Which of the curves in Fig. 37-15 best gives ∆t(vertical axis of the graph) versus speed parameterβ?

In Fig. 37-28a, particle P is to move parallel to thex andx' axes of reference framesSand S', at a certain velocity relative to frame S. FrameS' is to move parallel to thex axis of frame Sat velocity v. Figure 37-28b gives the velocityu' of the particle relative to frameS' for a range of values forv . The vertical axis scale is set byu'a=−0.800c . What value will u'have if (a) v=0.80cand (b)v→c ?

Question:A 2.50 MeV electron moves perpendicularly to a magnetic field in a path with a 3.0 cm radius of curvature. What is the magnetic field B? (See Problem 53)
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