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PHYSICS, M3 2023 VCE 12

A ray of monochromatic light is incident on a triangular glass prism with a refractive index of 1.52 . The ray is perpendicular to the side \(\text{AB}\) of the glass prism, as shown in Figure 14.
  

The ray of light travels through the glass prism before reaching side \(\text{AC}\).

  1. Calculate the critical angle for the glass prism at the glass-air interface.  (2 marks)

  2. Will the ray of light undergo total internal reflection at side \(\text{AC}\) of the glass prism? Justify your answer.  (2 marks)

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a.    41°

b.    As the angle of incidence 45° is greater than the critical angle 41°, the total internal reflection will occur on side \(\text{AC}\).

Show Worked Solution

a.     \(\sin \theta_c\) \(=\dfrac{n_2}{n_1}\)
  \(\theta_c\) \(=\sin^{-1}(\dfrac{n_2}{n_1})\)
    \(=\sin^{-1}(\dfrac{1.0}{1.52})\)
    \(=41^{\circ}\)

 

b.    Total internal reflection:

→ The angle of incidence 45° is greater than the critical angle 41°.

→ Therefore, total internal reflection will occur on side \(\text{AC}\).

♦ Mean mark (b) 46%.

PHYSICS, M3 2017 VCE 14

A light ray from a laser passes from a glucose solution \((n=1.44)\) into the air \((n=1.00)\), as shown in Figure 12.
 

  1. Calculate the critical angle (total internal reflection) from the glucose solution to the air.  (1 mark)

  2. The light ray strikes the surface at an angle of incidence to the normal of less than the critical angle calculated in part a.
  3. On Figure 12, sketch the ray or rays that should be observed.  (2 marks)

  4. The angle to the normal is increased to a value greater than the critical angle. An observer at point \(\text{X}\) in Figure 13 says she cannot see the laser.
     

  1. Explain why the observer says she cannot see the laser.  (2 marks)

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a.    \(\theta_c=44^{\circ}\)

b.    
       
→ As the light strikes the surface at an angle less than that of the critical angle, both reflection and refraction will be observed.
 
c.    Observer cannot see the laser:

→ When the light strikes the surface at an angle greater than the critical angle, Total Internal Reflection will occur.

→ As a result of the total internal reflection, all the light will be reflected off the boundary back into the glucose solution and therefore will not travel to the observer.

Show Worked Solution

a.     \(\sin\theta_c\) \(=\dfrac{n_2}{n_1}\)
  \(\theta_c\) \(=\sin^{-1}\Big{(}\dfrac{1.00}{1.44}\Big{)}\)
    \(=44^{\circ}\)

 

b.    
       
→ As the light strikes the surface at an angle less than that of the critical angle, both reflection and refraction will be observed.
♦♦ Mean mark (b) 37%.

c.    Observer cannot see the laser:

→ When the light strikes the surface at an angle greater than the critical angle, Total Internal Reflection will occur.

→ As a result of the total internal reflection, all the light will be reflected off the boundary back into the glucose solution and therefore will not travel to the observer.

PHYSICS, M3 2018 VCE 12

Optical fibres are constructed using transparent materials with different refractive indices.

Figure 14 shows one type of optical fibre that has a cylindrical core and surrounding cladding. Laser light of wavelength 565 nm is shone from air into the optical fibre (\(v=3 \times 10^8\)).
 

  1. Calculate the frequency of the laser light before it enters the optical fibre.  (1 mark)

  2. Calculate the critical angle for the laser light at the cladding-core boundary. Show your working.  (2 marks)

  3. Calculate the speed of the laser light once it enters the core of the optical fibre. Give your answer correct to three significant figures. Show your working.  (2 marks)

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a.    \(f=5.31 \times 10^{14}\ \text{Hz}\)

b.    \(\theta_c=60.3^{\circ}\)

c.    \(v_{\text{x}}=1.80 \times 10^8\ \text{ms}^{-1}\)

Show Worked Solution

a.     \(f\) \(=\dfrac{v}{\lambda}\)
    \(=\dfrac{3\times 10^8}{565 \times 10^{-9}}\)
    \(=5.31 \times 10^{14}\ \text{Hz}\)

 

b.     \(\theta_c\) \(=\sin^{-1}\Big{(}\dfrac{n_2}{n_1}\Big{)}\)
    \(=\sin^{-1} \Big{(}\dfrac{1.45}{1.67} \Big{)}\)
    \(=60.3^{\circ}\)

 

c.     \(v_{\text{x}}\) \(=\dfrac{c}{n_{\text{x}}}\)
    \(=\dfrac{3 \times 10^8}{1.67}\)
    \(=1.80 \times 10^8\ \text{ms}^{-1}\)
♦ Mean mark 50%.

PHYSICS, M3 2019 VCE 10*

The horizontal face of a glass block is covered with a film of liquid, as shown below.

A monochromatic light ray is incident on the glass-liquid boundary with an angle of incidence of 62.0°.
 

Calculate the minimum value of the liquid's refractive index, so that some light will just cross the interface into the liquid.   (2 marks)

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\(1.55\)

Show Worked Solution
\(\sin\theta_c\) \(=\dfrac{n_2}{n_1}\)  
\(n_2\) \(=\sin\theta_c \times n_1\)  
  \(= \sin62^{\circ} \times 1.75\)  
  \(=1.55\)  

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)

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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 12*

A 45° glass prism is submerged in water and oriented as shown in the diagram below. It is used to reflect a light ray through 90°.
 

What is the lowest refractive index of the glass prism that will ensure that total internal reflection occurs inside the prism?   (2 marks)

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\(1.88\)

Show Worked Solution

→ Using snell’s law where  \( \theta_2 = 90^{\circ}\).

\(n_1\sin\theta_1\) \(=n_2\sin\theta_2\)  
\(n_1 \times \sin 45^{\circ}\) \(=1.33 \times \sin 90^{\circ}\)  
\(n_1\) \(=\dfrac{1.33 \times 1}{\sin 45^{\circ}}\)  
  \(=1.88\)  
♦ Mean mark 51%.

PHYSICS, M3 2013 HSC 33b

Outline a first-hand investigation to demonstrate the transfer of light by optical fibres  (2 marks)

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→ Using a laser beam or other light source, shine the light into a glass rod at an angle at an angle greater than the critical angle.

→ The light beam can then be seen to reflect off either side of the glass rod as it travels down the line of the rod.

→ This is called total internal reflection (TIR) and demonstrates effectively how light travels through a optical fibre, As seen in the diagram below.
 

     

Show Worked Solution
 

→ Using a laser beam or other light source, shine the light into a glass rod at an angle at an angle greater than the critical angle.

→ The light beam can then be seen to reflect off either side of the glass rod as it travels down the line of the rod.

→ This is called total internal reflection (TIR) and demonstrates effectively how light travels through a optical fibre, As seen in the diagram below.