### 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}

$