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]
(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
(a) A transverse water wave travels to the right. The diagram shows the shape of the surface of the water at time \(t=0 . \mathrm{P}\) and \(\mathrm{Q}\) show two corks floating on the surface.
(i) State what is meant by a transverse wave.[1]
(ii) The frequency of the wave is \(0.50 \mathrm{~Hz}\). Calculate the speed of the wave.[1]
(iii) Plot on the diagram the position of \(\mathrm{P}\) at time \(t=0.50 \mathrm{~s}\).[1]
(iv) Show that the phase difference between the oscillations of the two corks is \(\pi\) radians.[1]
(b) Monochromatic light is incident on two very narrow slits. The light that passes through the slits is observed on a screen. \(M\) is directly opposite the midpoint of the slits. \(x\) represents the displacement from \(\mathrm{M}\) in the direction shown.
A student argues that what will be observed on the screen will be a total of two bright spots opposite the slits. Explain why the student’s argument is incorrect. [2]
(c) The graph shows the actual variation with displacement \(x\) from \(M\) of the intensity of the light on the screen. \(I_0\) is the intensity of light at the screen from one slit only.
(i) Explain why the intensity of light at \(x=0\) is \(4 I_0\).[2]
(ii) The slits are separated by a distance of \(0.18 \mathrm{~mm}\) and the distance to the screen is \(2.2 \mathrm{~m}\). Determine, in \(\mathrm{m}\), the wavelength of light.[2]
(iii) The two slits are replaced by many slits of the same separation. State one feature of the intensity pattern that will remain the same and one that will change.[2]
Stays the same:…………………………………………………………………………………………………….
Changes:……………………………………………………………………………………………………………….
(d) (i) Two sources are viewed though a single slit. The graph shows the diffraction pattern of one source.
Sketch, on the axes, the diffraction pattern of the second source when the images of the two sources are just resolved according to the Rayleigh criterion.[1]
(ii) Centaurus \(\mathrm{A}\) is a galaxy a distance of \(1.1 \times 10^{23} \mathrm{~m}\) away. A radio telescope of diameter \(300 \mathrm{~m}\) operating at a wavelength of \(3.2 \mathrm{~cm}\) is used to observe the galaxy. Determine the minimum size of the radio emitting region of the galaxy that can be resolved by this telescope.[2]
▶️Answer/Explanation
Ans:
a i «A wave where the displacement of particles/oscillations of particles/movement of particles/vibrations of particles is perpendicular/normal to the direction of energy transfer/wave travel/wave velocity/wave movement/wave propagation
a ii \(V=\alpha 0.50 \times 16=» 8.0 \ll \mathrm{ms}^{-1}\)
iv ALTERNATIVE 1
Phase difference is \(\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\) \(\alpha=\pi_b\)
ALTERNATIVE 2
One wavelength/period represents «phase difference» of \(2 \pi\) and «corks» are \(1 / 2\) wavelength/period apart so phase difference is \(\pi / O W T T E \checkmark\)
b.light acts as a wave «and not a particle in this situation» light at slits will diffract / create a diffraction pattern light passing through slits will interfere / create an interference pattern \&creating bright and dark spots»
c i The amplitude «at \(x=0\) will be doubled \(\checkmark\) intensity is proportional to amplitude squared / I \(\propto A^2 \checkmark\)
c ii Use of \(s=\frac{\lambda D}{d} \Rightarrow \lambda=\frac{s d}{D}\) OR \(s=\frac{n \lambda D}{d} \Rightarrow \lambda=\frac{s d}{n D}\)
$
\lambda=\ll \frac{0.567 \times 10^{-2} \times 0.18 \times 10^{-3}}{2.2}=» 4.6 \times 10^{-7} \ll \mathrm{mm}
$
iii Stays the same: Position/location of maxima/distance/separation between maxima «will be the same» / OWTTE \(\checkmark\)
Changes: Intensity/brightness/width/sharpness «of maxima will changew/ OWTTE \(\checkmark\)
d i . Maximum coinciding with first minimum \(A N D\) minimum coinciding with maximum \(\checkmark\)
d ii ALTERNATIVE 1
$
\begin{aligned}
& \frac{d}{D}=1.22 \times \frac{\lambda}{b} \text { therefore } d=\frac{1.22 \times \lambda \times D}{b} \\
& \varangle d \approx 1.22 \times \frac{3.2 \times 10^{-2} \times 1.1 \times 10^{23}}{300},=1.4 \times 10^{+9} \alpha \mathrm{m} *
\end{aligned}
$
ALTERNATIVE 2
$
\begin{aligned}
& \theta=\kappa 1.22 \frac{\lambda}{b}=1.22 \times \frac{3.2 \times 10^{-2}}{300}=» 1.3 \times 10^{-4} \text { «radians } » \\
& \mathrm{~d}=\alpha\left(1.1 \times 10^{23}\right)\left(1.3 \times 10^{-4}\right)=» 1.4 \times 10^{19} \ll \mathrm{m} »
\end{aligned}
$