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PHYSICS, M7 2020 VCE 12

In a Young's double-slit interference experiment, laser light is incident on two slits, \(\text{S}_1\) and \(\text{S}_2\), that are 4.0 × 10\(^{-4}\) m apart, as shown in Figure 11a.

Rays from the slits meet on a screen 2.00 m from the slits to produce an interference pattern. Point \(\text{C}\) is at the centre of the pattern. Figure 11b shows the pattern obtained on the screen.
 

  1. There is a bright fringe at point \(\text{P}\) on the screen.
  2. Explain how this bright fringe is formed.  (2 marks)
  1. The distance from the central bright fringe at point \(\text{C}\) to the bright fringe at point P is 1.26 × 10\(^{-2}\) m.
  2. Calculate the wavelength of the laser light. Show your working.  (3 marks

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a.    Bright fringe at point \(\text{P}\):

→ Young’s double slit experiment demonstrates that light can produce an interference pattern, from both constructive and destructive interference.

→ Point \(\text{P}\) is the 4th bright fringe and it follows that the path difference between the two light beams is 4 wavelengths.

→ At this point, the peaks of the twos wave constructively interfere and a bright band is formed.

b.    \(630\ \text{nm}\)

Show Worked Solution

a.    Bright fringe at point \(\text{P}\):

→ Young’s double slit experiment demonstrates that light can produce an interference pattern, from both constructive and destructive interference.

→ Point \(\text{P}\) is the 4th bright fringe and it follows that the path difference between the two light beams is 4 wavelengths.

→ At this point, the peaks of the twos wave constructively interfere and a bright band is formed.
 

♦ Mean mark (a) 45%.

b.    Using  \(d\,\sin \theta=m \lambda\)  and  \(\sin \theta=\dfrac{\Delta x}{D}\)

→ \(\Delta x\) is the distance between two bright bands and \(D\) is the distance from the slits to the screen.

\(\dfrac{d\Delta x}{D}\) \(=m\lambda\)  
\(\lambda\) \(=\dfrac{d\Delta x}{Dm}\)  
  \(=\dfrac{4 \times 10^{-4} \times 1.26 \times 10^{-2}}{2 \times 4}\)  
  \(=630\ \text{nm}\)  
♦ Mean mark (b) 40%.

PHYSICS, M7 2021 VCE 13

In Young's double-slit experiment, the distance between two slits, S\(_1\) and S\(_2\), is 2.0 mm. The slits are 1.0 m from a screen on which an interference pattern is observed, as shown in Figure 13a. Figure 13b shows the central maximum of the observed interference pattern.
 

  1. If a laser with a wavelength of 620 nm is used to illuminate the two slits, what would be the distance between two successive dark bands? Show your working.  (2 marks)
  1. Explain how this experiment supports the wave model of light.  (2 marks)

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a.    \(3.1 \times 10^{-4}\ \text{m}\)

b.    The experiment supports the wave theory as follows:

→ Young’s double slit experiment demonstrates how light will refract and form an interference pattern against a screen.

→ Since interference is a wave phenomenon, the experiment supports the wave model of light.

Show Worked Solution

a.    \(d\,\sin \theta=m \lambda\ \ \text{and}\ \ \sin \theta=\dfrac{\Delta x}{D}\)

→ \(\Delta x\) is the distance between two successive dark bands and \(D\) is the distance from the slits to the screen.

\(\dfrac{d\Delta x}{D}\) \(=m \lambda\)  
\(\Delta x\) \(=\dfrac{D\lambda}{d}\)  
  \(=\dfrac{1 \times 620 \times 10^{-9}}{2 \times 10^{-3}}\)  
  \(=3.1 \times 10^{-4}\ \text{m}\)  
♦ Mean mark (a) 51%.

 
b.    The experiment supports the wave theory as follows:

→ Young’s double slit experiment demonstrates how light will refract and form an interference pattern against a screen.

→ Since interference is a wave phenomenon, the experiment supports the wave model of light.

PHYSICS, M7 2023 VCE 13

A group of physics students undertake a Young's double-slit experiment using the apparatus shown in Figure 15. They use a green laser that produces light with a wavelength of 510 nm. The light is incident on two narrow slits, S\(_1\) and S\(_2\). The distance between the two slits is 100 \( \mu \)m.
 

An interference pattern is observed on a screen with points P\(_{0}\), P\(_{1}\) and P\(_2\) being the locations of adjacent bright bands, shown in Figure 15. Point  P\(_0\) is the central bright band.

  1. Calculate the path difference between S\(_{1}\)P\(_{2}\) and S\(_{2}\)P\(_{2}\). Give your answer in metres. Show your working.  (2 marks)
  1. The green laser is replaced by a red laser.
  2. Describe the effect of this change on the spacing between adjacent bright bands.  (1 mark)
  1. Explain how Young's double-slit experiment provides evidence for the wave-like nature of light and not the particle-like nature of light.  (3 marks)
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a.    \(1.02 \times 10^{-6}\ \text{m}\)

b.    The spacing between adjacent bright bands will increase.

c.    Evidence for the wave-like nature of light:

→ Young’s double slit experiment was used to show that light can produce an interference pattern as it diffracts when it passes through the slits.

→ As this is a wave phenomenon, the experiment provided valuable evidence to support the wave-like nature of light.

→ Using the particle-like nature of light as a model, two bright bands would have been expected behind the two slits as the particles would have passed through the slits in a straight line which was not observed.

Show Worked Solution

a.    \(\text{Difference in distance}\ = 2 \lambda\)

\(\text{Path difference}\ =2 \times 510 \times 10^{-9} = 1.02 \times 10^{-6}\ \text{m}\).

♦ Mean mark (a) 45%.

b.    The spacing will increase.

→ \( x=\dfrac{m\lambda D}{d}\), where \(D\) is the length between the screen and the slits and \(x\) is the distance between adjacent bright bands.

→ Since \(x \propto \lambda\), as the wavelength of light increases from green to red, so will the spacing between adjacent bright bands.
 

c.    Evidence for the wave-like nature of light:

→ Young’s double slit experiment was used to show that light can produce an interference pattern as it diffracts when it passes through the slits.

→ As this is a wave phenomenon, the experiment provided valuable evidence to support the wave-like nature of light.

→ Using the particle-like nature of light as a model, two bright bands would have been expected behind the two slits as the particles would have passed through the slits in a straight line which was not observed.

PHYSICS, M7 2023 VCE 18 MC

Which one of the following statements best describes the type of light produced from different types of light sources?

  1. Light from a laser is coherent and has a very narrow range of wavelengths.
  2. Light from an incandescent lamp is coherent and has a range of wavelengths.
  3. Light from an incandescent lamp is incoherent and has a very narrow range of wavelengths.
  4. Light from a single-colour light-emitting diode (LED) is coherent and contains a very wide range of wavelengths.
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\(A\)

Show Worked Solution

→ Laser light is coherent and usually monochromatic, meaning it produces a specific wavelength of light.

→ Incandescent light is incoherent but displays a range of wavelengths, therefore it is not \(B\) or \(C\).

→ LED’s contain a narrow range of wavelengths, therefore it is not \(D\).

\(\Rightarrow A\)

PHYSICS, M7 2023 HSC 3 MC

A diagram representing a double slit experiment using light is shown.
  

Which of the following best represents the expected pattern on the screen?
 

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\(C\)

Show Worked Solution

→ Light waves diffract when they pass through a double slit and produce an interference pattern.

→ The central maximum will occur in the centre of the screen where the path difference is 0 and where the path lengths differ by integral wavelengths.

\(\Rightarrow C\)

PHYSICS M7 2022 HSC 27

A laser producing red light of wavelength 655 nm is directed onto double slits separated by a distance, `d=5.0 xx 10^{-5} \ text{m}`. A screen is placed behind the double slits.
 


 

  1. Newton proposed a model of light. Use a labelled sketch to show the pattern on the screen that would be expected from Newton's proposed model.  (2 marks)
     

     
  2. When the laser light is turned on, a series of vertical bright lines are seen on the screen.
     
 
  1. Calculate the angle, `\theta`, between the centre line and the bright line at `A`.  (3 marks)
  1. The laser is replaced with one producing green light of wavelength 520 nm.
  2. Explain the difference in the pattern that would be produced.  (2 marks)
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a.   
       
 

b.   `theta=1.50^@`

c.   Consider  `d sin theta = m lambda :`

→ `sin theta prop lambda`

→ Using light with a shorter wavelength decreases the angular separation of bright fringes.

→ The bright lines will appear closer together.

Show Worked Solution
a.   
     

Mean mark part (a) 53%.

 

b.   Using  `d  sin  theta=mlambda:`

`5.0 xx10^(-5)  sin  theta` `=2 xx6.55xx10^(-7)`
   `sin  theta` `=(2 xx6.55xx10^(-7))/(5.0 xx10^(-5))`
   `theta` `=1.50^@`

 

c.   Consider  `d sin theta = m lambda :`

→ `sin theta prop lambda`

→ Using light with a shorter wavelength decreases the angular separation of bright fringes.

→ The bright lines will appear closer together.

PHYSICS, M7 2020 HSC 27

The following apparatus is used to investigate light interference using a double slit.
 

The distance, `y`, from the slits to the screen can be varied. The adjustment screws `(S)` vary the distance, `d`, between the slits. The wavelength of the laser light can be varied across the visible spectrum. The diffraction pattern shown is for a specific wavelength of light.

Explain TWO methods of keeping the distance between the maxima at `A` and `B` constant when the wavelength of the laser light is reduced.   (4 marks)

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`dsin theta=m lambda`

Decreasing the wavelength of light decreases the angular separation of the maxima.

To keep the distance between `A` and `B` constant:

→ Decrease the slit separation`d` causing the angular separation to increase, compensating for the effect of decreasing `lambda`. 

→ Increase the distance `y` between the slits and the screen. This increases the linear distance between `A` and `B`, mitigating the effect of the reduction in `lambda`.

Show Worked Solution

`dsin theta=m lambda`

Decreasing the wavelength of light decreases the angular separation of the maxima.

To keep the distance between `A` and `B` constant:

→ Decrease the slit separation`d` causing the angular separation to increase, compensating for the effect of decreasing `lambda`. 

→ Increase the distance `y` between the slits and the screen. This increases the linear distance between `A` and `B`, mitigating the effect of the reduction in `lambda`.


♦ Mean mark 46%.

PHYSICS, M7 2021 HSC 33

Two experiments are performed with identical light sources having a wavelength of 400 nm.

In experiment `A`, the light is incident on a pair of narrow slits `5.0 xx 10^(-5)` m apart, producing a pattern on a screen located 3.0 m behind the slits.
 

In experiment `B`, the light is incident on different metal samples inside an evacuated tube as shown. The kinetic energy of any emitted photoelectrons can be measured.
 

Some results from experiment `B` are shown.

How do the results from Experiment `A` and Experiment `B` support TWO different models of light? In your answer, include a quantitative analysis of each experiment.   (9 marks)

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→ Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.

→ When light is incident upon the slits, it diffracts and causes the slit to act as a source of wavefronts. When light from the slits arrives at the screen, bright bands are produced when light waves arrive in phase and undergo constructive interference.

→ Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.

→ The spacing between adjacent bright bands can be calculated:

              `d sin theta` `=m lambda`  
`5xx10^(-5)sin theta` `=1xx400 xx10^(-9)`  
                      `theta` `=0.46^(@)`  
                      `s` `=3 xx tan(0.46) = 0.024` m  

 
→ Experiment B supports Einstein’s particle, or photon model of light. This model can calculate the photon energy of incident light and explain why photons are emitted from calcium but not nickel:

`f=c/lambda =(3.00 xx10^(8))/(400 xx10^(-9))=7.50 xx10^(14)` Hz

`E=hf=6.626 xx10^(-34)xx7.50 xx10^(14)=4.97 xx10^(-19)` J
 

→ This energy is greater than the work function of calcium, explaining why one photon has enough energy to liberate a photoelectron from the calcium sample. However, this energy is less than the work function of nickel, explaining why no photoelectrons were observed from the nickel sample.

→ These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:

`K_(max)=hf-phi=4.97 xx10^(-19)-4.60 xx10^(-19)=3.70 xx10^(-20)\ \text{J}`

Show Worked Solution

→ Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.

→ When light is incident upon the slits, it diffracts and causes the slit to act as a source of wavefronts. When light from the slits arrives at the screen, bright bands are produced when light waves arrive in phase and undergo constructive interference.

→ Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.

→ The spacing between adjacent bright bands can be calculated:

              `d sin theta` `=m lambda`  
`5xx10^(-5)sin theta` `=1xx400 xx10^(-9)`  
                      `theta` `=0.46^(@)`  
                      `s` `=3 xx tan(0.46) = 0.024` m  

 
→ Experiment B supports Einstein’s particle, or photon model of light. This model can calculate the photon energy of incident light and explain why photons are emitted from calcium but not nickel:

`f=c/lambda =(3.00 xx10^(8))/(400 xx10^(-9))=7.50 xx10^(14)` Hz

`E=hf=6.626 xx10^(-34)xx7.50 xx10^(14)=4.97 xx10^(-19)` J
 

→ This energy is greater than the work function of calcium, explaining why one photon has enough energy to liberate a photoelectron from the calcium sample. However, this energy is less than the work function of nickel, explaining why no photoelectrons were observed from the nickel sample.

→ These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:

`K_(max)=hf-phi=4.97 xx10^(-19)-4.60 xx10^(-19)=3.70 xx10^(-20)\ \text{J}`


♦ Mean mark 52%.