IB Physics HL C.3 Wave Phenomena Question Bank
Practise IB Physics HL C.3 by solving quantitative interference, diffraction and refraction problems from complex wave data.
- Syllabus
- First assessment 2025
- Course
- Physics HL
- Level
- HL
Practise IB Physics HL C.3 by solving quantitative interference, diffraction and refraction problems from complex wave data.
A converging lens is placed between an object and a screen. An image of the object is formed on the screen.

Draw a ray to locate the focal point of the lens. Label this point with the letter F .
a

one of the two rays above
The lens suffers from spherical aberration.
Draw lines to complete the rays in the diagram.

i

the extreme ray crosses principal axis closer than paraxial ray
Monochromatic light enters the base of a plastic beaker that contains water with an oil layer floating on it. A student draws a diagram to show the directions the light takes as it passes through the layers. The student's diagram has one error at position A and one error at position B. The refractive indices of the materials are shown on the diagram.

The light is refracted at an angle of 32∘ when it enters the plastic layer as shown.
Identify, with a reason, the error in the student's diagram for
light crossing the plastic-water interface (position A).
the angle of refraction ought to be greater than the angle of incidence OR the ray should refract away from the normal
because ray goes from high refractive index/<<optically>> more dense/slower medium to
low refractive index/optically less dense/faster medium
Marking guidance:
Do not allow use of e.g n1
unless medium one is
described e.g. nair
light at the water-oil interface (position B).
there should be a <<transmitted>> ray in the oil
OR
total internal reflection is not possible
because ray goes from low refractive index/<<optically>> less dense/faster medium to high refractive index/<<optically>> more dense/slower medium
Calculate the angle of incidence at the air-plastic interface.
Use of Snell's Law « sinrsini=11.60 »
i=sin−1≪1.60×sin32∘>=58<∘≫
'Use of' requires a substitution NOT just a statement of a formula
Marking guidance:
Accept 1.0 rad (unit must be included to show a deliberate attempt to use rad rather than a calculator mistake)
Calculate the critical angle for the plastic-water interface.
sinrsini=1.601.33 and sinr=1i= «sin −10.831»=56 " ∘ »
Accept 0.98 rad (unit required)
Monochromatic light of wavelength 6.3×10−7 m in air is incident from above at a normal to the oil layer. Rays on the diagram are shown at near-normal incidence for clarity. Three positions X, Y and Z are shown on the diagram.

Identify, with a reason, a position at which there is a phase change of 180∘.
position X because light reflects off the medium of higher refractive index
Marking guidance:
Allow correct references to optical density or speed as in previous questions
A statement of X and that the refractive index of oil is greater than the refractive index of air is sufficient
Determine the minimum thickness of the oil layer for which light is not reflected. State your answer to an appropriate number of significant figures.
Use of 2dn=mλ2.1×10−7 «m»
any answer to 2 s.f.
'Use of' requires a substitution
NOT just a statement of a formula
A beam of coherent monochromatic light from a distant galaxy is used in an optics experiment on Earth.
The beam is incident normally on a double slit. The distance between the slits is 0.300 mm . A screen is at a distance D from the slits. The diffraction angle θ is labelled.

A series of dark and bright fringes appears on the screen. Explain how a dark fringe is formed.
superposition of light from each slit / interference of light from both slits with path/phase difference of any half-odd multiple of wavelength/any odd multiple of π (in words or symbols) producing destructive interference
Marking guidance:
Ignore any reference to crests and troughs.
Outline why the beam has to be coherent in order for the fringes to be visible.
light waves (from slits) must have constant phase difference / no phase difference / be in phase
OWTTE
The wavelength of the beam as observed on Earth is 633.0 nm . The separation between a dark and a bright fringe on the screen is 4.50 mm . Calculate D.
evidence of solving for D≪D=λsd>∨<633.0×10−94.50×10−3×0.300×10−3×2>=4.27<m≫
Marking guidance:
Award [1] max for 2.13 m.
The graph of variation of intensity with diffraction angle for this experiment is shown.

Calculate the angular separation between the central peak and the missing peak in the double-slit interference intensity pattern. State your answer to an appropriate number of significant figures.
sinθ=0.300×10−34×633.0×10−9θ=0.0084401…
final answer to three sig figs (eg 0.00844 or 8.44×10−3 )
Allow ECF from (a)(iii).
Award [1] for 0.121 rad (can award MP3 in addition for proper sig fig)
Accept calculation in degrees leading to 0.481 degrees.
Award MP3 for any answer expressed to 3 s f.
The air between the slits and the screen is replaced with water. The refractive index of water is 1.33 .
Calculate the wavelength of the light in water.
1.33633.0=476 « nm »
State two ways in which the intensity pattern on the screen changes.
distance between peaks decreases intensity decreases