IBDP Physics-C.5 Doppler effect- IB Style Questions For HL Paper 1A -FA 2025

Question

A source moving with speed $v$ away from a stationary observer emits light of wavelength $\lambda$. The wavelength received by the observer is $\lambda+\Delta\lambda$. The speed $v$ is much less than the speed of light.

Which graph gives the variation of $\Delta\lambda$ with $v$?

▶️ Answer/Explanation

Correct Answer: $ \boxed{\mathrm{D}} $

For a source moving away from a stationary observer, the relativistic Doppler effect gives

$\lambda_{\mathrm{obs}}=\lambda\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}$

Since $v\ll c$, the expression can be approximated by

$\lambda_{\mathrm{obs}}\approx\lambda\left(1+\dfrac{v}{c}\right)$

Therefore,

$\lambda+\Delta\lambda\approx\lambda+\dfrac{\lambda v}{c}$

Hence,

$\Delta\lambda\approx\dfrac{\lambda}{c}v$

Thus, $\Delta\lambda$ is directly proportional to $v$. The graph must therefore be a straight line passing through the origin.

Hence, the correct answer is $ \boxed{\mathrm{D}} $.

Question

A source $S$ of a sound wave is moving at a constant velocity along a line joining stationary observers $X$ and $Y$. The diagram shows consecutive wavefronts of the sound.

Consider the following statements:

I. More wavefronts pass $X$ in a unit time than $Y$.

II. The speed of the wavefronts is greater at $X$ than at $Y$.

III. The wavelength of the sound at $X$ is less than the wavelength at $Y$.

Which statements are correct?

(A) I and II only
(B) I and III only
(C) II and III only
(D) I, II and III

▶️ Answer/Explanation

Correct Answer: $ \boxed{\mathrm{B}} $

The source $S$ is moving towards observer $X$ and away from observer $Y$. Therefore, the wavefronts are compressed in front of the moving source and spread out behind it.

Statement I: More wavefronts pass $X$ in a unit time than $Y$.

This is correct. The wavefronts are closer together at $X$, so the observed frequency at $X$ is greater than at $Y$.

Since frequency is the number of wavefronts passing a point per unit time, more wavefronts pass $X$ in a given time.

Statement II: The speed of the wavefronts is greater at $X$ than at $Y$.

This is incorrect. The speed of sound in a given medium is determined by the properties of the medium and is independent of the motion of the source. Therefore, the wavefront speed is the same at $X$ and $Y$.

Statement III: The wavelength of the sound at $X$ is less than the wavelength at $Y$.

This is correct. The wavefronts are compressed in front of the source, so

$\lambda_X<\lambda_Y$

The relationship

$v=f\lambda$

also shows that, since the wave speed $v$ is constant, the higher frequency at $X$ corresponds to a shorter wavelength.

Therefore, statements I and III are correct, while statement II is incorrect.

Hence, the correct answer is $ \boxed{\mathrm{B}} $.

Question

A radar detector is used to measure the speed of a car. The car is moving with a speed v towards the detector.

The detector emits microwaves of frequency f and speed c. Which of the following is the change in frequency of the microwaves measured at the detector after reflection by the car?

▶️Answer/Explanation

Ans C

(First Shift)

The car moves toward the radar detector with speed $ v $, and the radar emits microwaves of frequency $ f $ and speed $ c $.
The frequency received by the moving car is given by the Doppler formula

$
f’ = f \left(\frac{c + v}{c}\right)
$

where:
$ f’ $ is the frequency detected by the moving car.
$ c $ is the speed of light.

Since the car is moving toward the source, the frequency increases.

(Second Shift)
The moving car acts as a new source, reflecting the frequency $ f’ $ back to the detector.
the detector is stationary and the car is moving toward it, another Doppler shift occurs

$
f” = f’ \left(\frac{c + v}{c}\right)
$

$
f” = f \left(\frac{c + v}{c}\right) \times \left(\frac{c + v}{c}\right)
$

$
f” = f \left(\frac{(c + v)^2}{c^2}\right)
$

Expanding for small $ v/c $, using approximation $ (1 + x)^2 \approx 1 + 2x $ for small $ x $

$
f” \approx f \left(1 + \frac{2v}{c}\right)
$

$
\Delta f = f” – f = f \frac{2v}{c}
$

IBDP Physics-C.5 Doppler effect- IB Style Questions For HL Paper 1A -FA 2025

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