# IBDP Physics C.5 Doppler effect IB Style Question Bank : HL Paper 2

### Question

A girl looks at a flat vertical glass window on an automobile. The window has a thin transparent coating.

White light from a source is reflected to the girl from the coating and from the window. (a) Outline why the light reflected to the girl has one wavelength missing. 

(b) The refractive index of the coating is 1.63 and the refractive index of the glass is 1.52 .
The thickness of the coating is $$143 \mathrm{~nm}$$.
Determine the wavelength, in $$\mathrm{nm}$$, that is missing in the light reflected to the girl assuming that the light is incident normally on the window.

(b) The refractive index of the coating is 1.63 and the refractive index of the glass is 1.52 . The thickness of the coating is $$143 \mathrm{~nm}$$.
Determine the wavelength, in $$\mathrm{nm}$$, that is missing in the light reflected to the girl assuming that the light is incident normally on the window. 

(c) The automobile is driven directly away from the girl at a steady speed of $$15 \mathrm{~m} \mathrm{~s}^{-1}$$. A musical note is emitted by a loudspeaker in the automobile.
The frequency of the musical note heard by the girl is $$410 \mathrm{~Hz}$$.
(i) Outline why the driver of the automobile and the girl hear different frequencies for the musical note.
(ii) The speed of sound in air is $$330 \mathrm{~ms}^{-1}$$.
Calculate the frequency of the musical note emitted by the loudspeaker.

Ans:

There is «partialm reflection at the front surface of the layer and also at the glasslayer interface $$\checkmark$$
destructive interference «between these reflections occurs at the missing wavelength

b Use of $$2 d n=m \lambda \checkmark$$
Use of $$n=1.63$$
470 «nm»

c i. There is relative motion between the girl and the automobile «Relative to the girl» the wavelength of the sound changes/increases $$\checkmark$$

c ii. Use of $$f=f^{\prime}\left(\frac{c+u}{c}\right)$$. $$\alpha f=410\left(\frac{330+15}{330}\right)=430 \alpha \mathrm{Hz}$$

### Question

Sound of frequency f = 2500 Hz is emitted from an aircraft that moves with speed v = 280 ms–1 away from a stationary observer.

The speed of sound in still air is c = 340 ms–1. 1. frequency heard by the observer. 

2. wavelength measured by the observer. 

f’ = 2500 × $$\frac{340}{340+280}$$
ii λ = $$\frac{340}{1371}$$ ≈ 0.24 / 0.25 «m »