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. [2]
(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.[3]
(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. [3]
(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.[2]
(ii) The speed of sound in air is \(330 \mathrm{~ms}^{-1}\).
Calculate the frequency of the musical note emitted by the loudspeaker.[2]
▶️Answer/Explanation
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.
frequency heard by the observer. [2]
wavelength measured by the observer. [1]
Answer/Explanation
Ans
i
f’ = 2500 × \(\frac{340}{340+280}\)
f ′ = 1371 ≈1400 «Hz »
ii λ = \(\frac{340}{1371}\) ≈ 0.24 / 0.25 «m »