Chapter 6: Q23E (page 1249)

(a) How fast must you be approaching a red traffic light $(λ = 675 nm)$ for it to appear yellow $(A = 575 nm)$ ? Express your answer in terms of the speed of light. (b) If you used this as a reason not to get a ticket for running a red light, how much of a fine would you get for speeding? Assume that the fine is $$ 1.00$ for each kilometer per hour that your speed exceeds the posted limit of $90 km / h$ .

Short Answer

Expert verified

The red traffic light moving with 0.159c speed of light and the fine is $1.72 \times 10^{8} milliondollar$ .

Step by step solution

Definition of frequency

The term frequency may be defined as the number of times something happens in a particular time period.

Determine the red-light moving speed and price of ticket.

The moving speed of red light can be calculated by the relation

$u = c (\frac{f^{2} - f_{0}^{2}}{f_{0}^{2} + f^{2}})$

Now solving for $f, f_{0}$

$f = \frac{c}{λ_{yellow}} f = \frac{2.998 \times 10^{8} m / s}{575 \times 10^{- 9} m} f = 5.21 \times 10^{14} Hz f = \frac{c}{λ_{yellow}} f_{0} = \frac{2.998 \times 10^{8} m / s}{675 \times 10^{- 9} m} f_{0} = 4.44 \times 10^{- 9} Hz$

Now put both values in above result than

$u = c (\frac{{(5.21 \times 19^{14} H z)}^{2} - {(4.44 \times 10^{14} H z)}^{2}}{{(4.44 \times 10^{14} H z)}^{2} - {(5.21 \times 19^{14} H z)}^{2}}) u = 0.159 c$

Hence, the red traffic light moving with 0.159c speed of light.

Now for the ticket price

$0.159 c = 0.159 (3 \times 10^{5} km / s) 0.159 c = (4.77 \times 10^{4} km / s) (\frac{3600 s}{1 h}) 0.159 c = 1.72 \times 10^{8} Km / h$

Hence, the fine is $$ 1.72 \times 10^{8}$ dollar.

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Short Answer

Step by step solution

Definition of frequency

Determine the red-light moving speed and price of ticket.

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