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PHYSICS, M3 2021 VCE 12

A Physics teacher is conducting a demonstration involving the transmission of light within an optical fibre. The optical fibre consists of an inner transparent core with a refractive index of 1.46 and an outer transparent cladding with a refractive index of 1.42. A single monochromatic light ray is incident on the optical fibre, as shown in Figure 12.
 

  1. Determine the angle of incidence, \(\theta\), at the air-core boundary. Show your working.  (2 marks)

  2. Will any of the initial light ray be transmitted into the cladding? Explain your answer and show any supporting working.  (3 marks)

Show Answers Only

a.   \(\theta=51^{\circ}\)

b.    Find the critical angle for Total Internal Reflection (TIR):

\(\sin\theta_c\) \(=\dfrac{n_2}{n_1}\)  
\(\theta_c\) \(=\sin^{-1}\Big{(}\dfrac{n_2}{n_1}\Big{)} \)  
  \(=\sin^{-1} \Big{(}\dfrac{1.42}{1.46} \Big{)}\)  
  \(=76.6^{\circ}\)  

 
→ The angle of incidence is  \(90^{\circ}-32^{\circ}=58^{\circ}\).

→ As the angle of incidence is less than the critical angle, TIR will not occur and light will be transmitted into the cladding.

Show Worked Solution

a.    Using Snell’s Law:

\(n_1\sin\theta_1\) \(=n_2\sin\theta_2\)  
\(\theta_1\) \(=\sin^{-1} \Big{(}\dfrac{n_2\sin\theta_2}{n_1} \Big{)} \)  
  \(=\sin^{-1}\Big{(}\dfrac{1.46 \times \sin32^{\circ}}{1.0}\Big{)} \)  
  \(=51^{\circ}\)  

 
b.    Find the critical angle for Total Internal Reflection (TIR):

\(\sin\theta_c\) \(=\dfrac{n_2}{n_1}\)  
\(\theta_c\) \(=\sin^{-1}\Big{(}\dfrac{n_2}{n_1}\Big{)} \)  
  \(=\sin^{-1} \Big{(}\dfrac{1.42}{1.46} \Big{)}\)  
  \(=76.6^{\circ}\)  

 
→ The angle of incidence is  \(90^{\circ}-32^{\circ}=58^{\circ}\).

→ As the angle of incidence is less than the critical angle, TIR will not occur and light will be transmitted into the cladding.

♦ Mean mark (b) 45%.

PHYSICS, M3 2022 VCE 13

A ray of green light from a light-emitting diode (LED) strikes the surface of a tank of water at an angle of 40.00° to the surface of the water, as shown in Figure 11. The ray arrives at the base of the tank at point \(\text{X}\). The depth of the water in the tank is 80.00 cm. The refractive index of green LED light in water is 1.335
 

  1. Calculate the distance \(\text{OX}\). Outline your reasoning and show all your working. Give your answer in centimetres.  (4 marks)

  2. The green LED light is replaced with a narrow beam of white sunlight.
  3. Describe the colour of the light that arrives to the left of point \(\text{X}\), at point \(\text{X}\) and to the right of point \(\text{X}\).  (3 marks)
     
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a.  \(OX=56\ \text{cm}\)

b.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Light to the left of point X}\rule[-1ex]{0pt}{0pt} & \text{Light at point X} & \text{Light to the right of point X} \\
\hline
\rule{0pt}{2.5ex}\text{Blue\Purple}\rule[-1ex]{0pt}{0pt} & \text{Green} & \text{Red} \\ \text{(lower wavelength)} & & \text{(higher wavelength)} \\
\hline
\end{array}

Show Worked Solution

a.  \(\text{Using Snell’s Law:}\)

\(n_1 \sin \theta_1\) \(=n_2 \sin \theta_2\)  
\(\sin \theta_2\) \(=\dfrac{n_1 \sin \theta_1}{n_2}\)  
\( \theta_2\) \(= \sin^{-1}\Big{(}\dfrac{1 \times \sin 50^{\circ}}{1.335} \Big{)} \)  
  \(=35^{\circ}\)  

 

\(\tan35^{\circ}\) \(=\dfrac{OX}{80}\)  
\(OX\) \(=80\times \tan35^{\circ}\)  
  \(=56\ \text{cm}\)

♦ Mean mark 48%.

b. 

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Light to the left of point X}\rule[-1ex]{0pt}{0pt} & \text{Light at point X} & \text{Light to the right of point X} \\
\hline
\rule{0pt}{2.5ex}\text{Blue\Purple}\rule[-1ex]{0pt}{0pt} & \text{Green} & \text{Red} \\ \text{(lower wavelength)} & & \text{(higher wavelength)} \\
\hline
\end{array}