Waves Practice Problems
Waves and optics practice problems on wave speed, refraction, lenses, photon energy and decibels, marked to within 1.5 percent with the working shown.
Practice
Question 1 of 40
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Worked answers
The first ten questions from the set above, each with its answer and the working that gets there. The working is carried out in the units each equation takes, so its last line can show the answer before it is converted.
-
- Frequency
- f = 1700 THz
- Wave speed
- v = 2.998 × 10⁸ m/s
Find the wavelength (λ).
Show the answer and working
Answer λ = 176.3 nm
Rearranged
λ = v ÷ f- lambda = v / f
- = 2.99792 × 10⁸ m/s / 1700 THz
- = 2.99792 × 10⁸ / (1.7 × 10¹⁵)
- = 1.763 × 10⁻⁷ m = 176.35 nm
Check it with the Wavelength and Frequency Calculator.
-
- Photon energy
- E = 8.5 eV
Find the wavelength (λ).
Show the answer and working
Answer λ = 145.9 nm
Rearranged
λ = hc ÷ E- lambda = h c / E
- h c = 1.9864 × 10⁻²⁵ J m
- = 1.9864 × 10⁻²⁵ / (1.362 × 10⁻¹⁸ J)
- = 1.459 × 10⁻⁷ m = 145.86 nm
Check it with the Photon Energy Calculator.
-
- Refractive index, medium 1
- n₁ = 2.2
- Angle of incidence
- θ₁ = 32°
- Refractive index, medium 2
- n₂ = 1.3
Find the angle of refraction (θ₂).
Show the answer and working
Answer θ₂ = 63.74°
Rearranged
θ₂ = arcsin(n₁ sin θ₁ / n₂)- n1 sin(t1) = n2 sin(t2)
- sin(t2) = 2.2 x sin(32 deg) / 1.3
- sin(t2) = 0.89679
- t2 = 63.74 deg
Check it with the Snell’s Law Calculator.
-
- Focal length
- f = 16 cm
- Image distance
- dᵢ = 37 cm
Find the object distance (dₒ).
Show the answer and working
Answer dₒ = 28.19 cm
Rearranged
1/dₒ = 1/f − 1/dᵢ- 1/f = 1/do + 1/di
- 1/0.16 = 1/0.2819 + 1/0.37
Check it with the Thin Lens Calculator.
-
- Sound level
- L = 22 dB
- Reference intensity
- I₀ = 1 pW/m²
Find the intensity (I).
Show the answer and working
Answer I = 0.0001585 µW/m²
Rearranged
I = I₀ × 10^(L ÷ 10)- I = I0 x 10^(L / 10)
- = 1 × 10⁻¹² x 10^(22 / 10)
- = 1 × 10⁻¹² x 10^2.2
- = 1.585 × 10⁻¹⁰ W/m2 = 0.00015849 uW/m2
Check it with the Decibel Calculator.
-
- Wavelength
- λ = 2100 nm
- Wave speed
- v = 2.998 × 10⁸ m/s
Find the frequency (f).
Show the answer and working
Answer f = 142.8 THz
Rearranged
f = v ÷ λ- f = v / lambda
- = 2.99792 × 10⁸ m/s / 2.1 um
- = 2.99792 × 10⁸ / (2.1 × 10⁻⁶ m)
- = 1.4276 × 10¹⁴ Hz = 142.76 THz
Check it with the Wavelength and Frequency Calculator.
-
- Wavelength
- λ = 1100 nm
Find the photon energy (E).
Show the answer and working
Answer E = 1.127 eV
Rearranged
E = hc ÷ λ- E = h c / lambda
- h c = 6.62607 × 10⁻³⁴ x 2.99792 × 10⁸ = 1.9864 × 10⁻²⁵ J m
- = 1.9864 × 10⁻²⁵ / (1.1 × 10⁻⁶ m)
- = 1.806 × 10⁻¹⁹ J = 1.1271 eV
Check it with the Photon Energy Calculator.
-
- Refractive index, medium 1
- n₁ = 2
- Refractive index, medium 2
- n₂ = 1.5
- Angle of refraction
- θ₂ = 81°
Find the angle of incidence (θ₁).
Show the answer and working
Answer θ₁ = 47.8°
Rearranged
θ₁ = arcsin(n₂ sin θ₂ / n₁)- n1 sin(t1) = n2 sin(t2)
- 2 x sin(47.8) = 1.5 x sin(81)
Check it with the Snell’s Law Calculator.
-
- Wavelength
- λ = 230 nm
- Wave speed
- v = 2.998 × 10⁸ m/s
Find the frequency (f).
Show the answer and working
Answer f = 1303 THz
Rearranged
f = v ÷ λ- f = v / lambda
- = 2.99792 × 10⁸ m/s / 230 nm
- = 2.99792 × 10⁸ / (2.3 × 10⁻⁷ m)
- = 1.3034 × 10¹⁵ Hz = 1303.4 THz
Check it with the Wavelength and Frequency Calculator.
-
- Photon energy
- E = 3.9 eV
Find the wavelength (λ).
Show the answer and working
Answer λ = 317.9 nm
Rearranged
λ = hc ÷ E- lambda = h c / E
- h c = 1.9864 × 10⁻²⁵ J m
- = 1.9864 × 10⁻²⁵ / (6.248 × 10⁻¹⁹ J)
- = 3.179 × 10⁻⁷ m = 317.91 nm
Check it with the Photon Energy Calculator.
Fix the geometry before the algebra
Optics problems are geometry with a formula attached, so draw the surface and the normal before writing anything down. Every angle in refraction is measured from the normal, the line perpendicular to the surface, so a ray striking glass at 20 degrees to the surface has an angle of incidence of 70 degrees. Note which medium is the denser one, because that decides which way the ray bends.
Then convert. Wavelengths are quoted in nanometres and the formulas want metres, so
550 nm is 5.5 × 10⁻⁷ m, and a terahertz is 10¹² hertz. For
lenses, settle the sign convention first: a converging lens has a positive focal length
and a positive image distance means a real image.
The conventions that cost marks
- Angles taken from the surface. The single most common error in refraction, and it produces an answer that still looks like an angle.
- The ratio of angles instead of the ratio of sines. Light entering
water at 30 degrees refracts to 22.03 degrees, from
1 × sin 30 = 1.333 × sin θ₂. Dividing 30 by 1.333 gives 22.5, close enough to pass unnoticed. - Decibels added. Two 60 dB sources together make 63 dB, not 120, because doubling the intensity adds 3.01 dB. Intensities are what add.
- Nanometres left in place. A wavelength entered as 550 rather than
5.5 × 10⁻⁷puts the frequency out by10⁹, which lands a green photon in the radio band. - A negative image distance read as an error. It is the answer: the image is virtual, on the same side as the object, which is what happens when the object sits inside the focal length.
Two checks that catch most slips
For photons, E = 1239.84 / λ gives the energy in electronvolts from a
wavelength in nanometres. A 550 nm green photon is 2.25 eV, and the visible range spans
about 1.59 eV at 780 nm to 3.26 eV at 380 nm. The wave equation pairs with it: 550 nm in
vacuum is 545 THz.
For refraction, direction is the check: light entering a denser medium always bends toward the normal, so the refracted angle is the smaller one. A critical angle exists only going the other way, at 48.6 degrees for water into air, 41.1 degrees for crown glass at 1.52 and 24.4 degrees for diamond. Sound levels scale in tens: a factor of ten in intensity is 10 dB, and doubling the distance from a point source costs 6.02 dB.
What changes at a boundary and what does not
When light crosses into a denser medium its speed and wavelength both drop by the factor
n, while its frequency does not change, because frequency is
set by the source. Light at 550 nm in vacuum is 413 nm inside water, still 545 THz, and
still green.
A decibel level is a ratio rather than a quantity, referenced for airborne sound to the
threshold of hearing at
1 × 10⁻¹² W/m², so 60 dB means a million times that intensity. And the speed
of sound depends on the medium and its temperature, about 343 m/s in air at 20 °C, which
makes a 440 Hz tone 0.78 m long.
Common questions
Do I have to match the number of significant figures?
No. An answer counts as correct within 1.5 percent of the computed value, so three significant figures carried through the working is always enough, and rounding an intermediate step will not fail you. The tolerance is relative rather than absolute because answers on this topic run from picowatts per square metre to hundreds of terahertz.
Are the angles measured from the normal or from the surface?
From the normal, the line perpendicular to the surface, which is the standard convention and the one the calculators use. So an angle of incidence of 70 degrees describes a ray at 20 degrees to the surface itself, close to grazing. Answers are marked against the same convention, which means an angle measured from the surface will be marked wrong even when the arithmetic behind it was sound.