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DP IB Physics: HL

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Home / IB / Physics: HL / DP / Topic Questions / 9. Wave Phenomena (HL only) / 9.4 Resolution / Multiple Choice


9.4 Resolution

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Question 1

Marks: 1

Which statement is not true about circular apertures?

  • Circular apertures allow a cone of light to enter a region behind the aperture

  • Circular apertures allow light to act like a point source

  • The diffraction of light through a circular aperture produces a diffuse disc surrounded by fainter concentric circular rings

  • The diffraction pattern produced by circular apertures produces a central linear fringe with subsequent alternating dark and light fringes

Choose your answer
  
Key Concepts
Diffracting Aperture

Question 2

Marks: 1

What does the Rayleigh criterion describe?

  • The diffraction of light through a single slit

  • The diffraction of light through a double slit

  • The diffraction of light through multiple slits

  • The minimum separation between two light sources that can be resolved into two distinct objects

Choose your answer
  
Key Concepts
The Rayleigh Criterion

Question 3

Marks: 1

The diagram below shows the light from two point sources, s1 and s2 passing through a circular aperture.

cfSbxKUZ_9-4-ib-hl-emcq-3-q-stem

What name is given to the angle shown as θ?

  • Angular separation

  • Angle of diffraction

  • Angle of incidence

  • Angle of reflection

Choose your answer
  
Key Concepts
Diffracting Aperture

Question 4

Marks: 1

Light from two sources with wavelength λ is passed through a circular aperture with diameter b, producing an angle of diffraction θ.

According to the Rayleigh criterion, how would the angle of diffraction be affected if the diameter of the aperture was doubled?

  • 0.25 space theta

  • 0.50 space theta

  • theta

  • 2 space theta

Choose your answer
  
Key Concepts
Rayleigh Criterion Calculations

Question 5

Marks: 1

Which of the following options would not increase the resolution of two objects?

  • Decreasing the wavelength of the light

  • Increasing the angular separation of the objects

  • Increasing the diameter of the aperture

  • Increasing the angle of diffraction

Choose your answer
  
Key Concepts
Rayleigh Criterion Calculations

Question 6

Marks: 1

The Rayleigh criterion can be presented as:

 s over d space greater or equal than space 1.22 lambda over b

Which option provides the correct quantities represented in the expression?

 

 

s

d

λ

b

A.

Distance between sources

Distance between source and aperture

Wavelength

Diameter of aperture

B.

Distance between source and aperture

Distance between sources

Wavelength

Diameter of aperture

C.

 Wavelength Diameter of aperture

Distance between sources

 		 Distance between source and aperture

D.

 Diameter of aperture

Wavelength

Distance between source and aperture

Distance between sources
    Choose your answer
      
    Key Concepts
    Rayleigh Criterion Calculations

    Question 7

    Marks: 1

    The Rayleigh criterion can also be applied to diffraction gratings. 

    Incident light with an average wavelength of 600 nm is incident on a diffraction grating. The difference between the wavelengths is 25 nm. How may slits would the grating need to have to resolve fully the second order of diffraction?

    • 24

    • 25

    • 12

    • 50

    Choose your answer
      
    Key Concepts
    Resolvance of Diffraction Gratings

    Question 8

    Marks: 1

    Which of the following objects can be resolved according to the Rayleigh criterion?

    9-4-ib-easy-mcq-q8

      Choose your answer
        
      Key Concepts
      The Rayleigh Criterion

      Question 9

      Marks: 1

      The Rayleigh criterion can be expressed as:

       s over d space greater or equal than space 1.22 lambda over b

      A telescope is aimed at two objects that are known to be 10 cm apart. The angle of diffraction is 4.5 × 10−5 rad.

      Can the objects be resolved at a distance of 10 km?

      • Yes, because 1 cross times 10 to the power of negative 5 end exponent space greater than space 4.5 cross times 10 to the power of negative 5 end exponent

      • No, because 1 cross times 10 to the power of negative 5 end exponent space less than space 4.5 cross times 10 to the power of negative 5 end exponent

      • Yes, because 1 cross times 10 to the power of negative 3 end exponent space greater than space 4.5 cross times 10 to the power of negative 5 to the power of blank end exponent

      • No, because 1 cross times 10 to the power of negative 3 end exponent space less than space 4.5 cross times 10 to the power of negative 5 end exponent

      Choose your answer
        
      Key Concepts
      Rayleigh Criterion Calculations

      Question 10

      Marks: 1

      It is not possible to view an object smaller than the wavelength used to see it. 

      Which of the following can be used to resolve objects smaller than the wavelengths of visible light, such as atoms?

      • Sodium ions

      • Electrons

      • Gold atoms

      • Water molecules

      Choose your answer
        
      Key Concepts
      Diffracting Aperture


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