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Students usually remember that light slows down in glass. Fewer can say what happens to its wavelength and frequency, and exam questions are built on exactly that gap. Here are the rules, why frequency is the one that stays fixed, and three worked examples using values from the NCERT table of refractive indices.
Where is this in the syllabus?
The NTA Syllabus for JEE (Main) 2026 lists refraction at plane and spherical surfaces and "laws of reflection and refraction using Huygens principle" in Unit 16, and the NEET (UG) 2026 syllabus has the same items. The NCERT Class 12 chapter Wave Optics (Reprint 2026-27) derives λ₁/λ₂ = v₁/v₂ and states that when a wave enters a denser medium "the wavelength and the speed of propagation decrease but the frequency ... remains the same." The Ray Optics chapter gives the refractive indices used below: water 1.33, crown glass 1.52.
What are the three rules?
For light going from vacuum (or air, to a good approximation) into a medium of refractive index n:
Speed: v = c/n
Wavelength: λ_medium = λ/n
Frequency: unchanged
The CODATA value of c is exactly 299 792 458 m/s. Our post on Huygens' principle shows the wavefront construction behind these rules.
Why does frequency stay the same?
Frequency is set by the source. At the boundary, every crest that arrives from the air must continue into the glass; crests cannot pile up at the surface or vanish there. So the number of crests passing per second, the frequency, is the same on both sides. Since v = fλ and v falls, λ must fall by the same factor.
Worked example 1: red light into crown glass
Light of wavelength 633 nm in air enters crown glass, n = 1.52.
Speed: 2.998 × 10⁸ / 1.52 = 1.97 × 10⁸ m/s
Wavelength: 633 / 1.52 = 416 nm
Frequency: 2.998 × 10⁸ / (633 × 10⁻⁹) = 4.74 × 10¹⁴ Hz, in air and in glass
Check inside the glass: 1.97 × 10⁸ / (416 × 10⁻⁹) = 4.74 × 10¹⁴ Hz
Worked example 2: a glass slab 2.0 mm thick
The same light crosses a crown glass plate 2.0 mm thick.
Wavelengths that fit in the plate: 2.0 × 10⁻³ / 416 × 10⁻⁹ ≈ 4800. In the same 2.0 mm of air: about 3160.
Time to cross: 2.0 × 10⁻³ / 1.97 × 10⁸ = 1.01 × 10⁻¹¹ s, against 6.67 × 10⁻¹² s in air. The delay is about 3.5 × 10⁻¹² s.
Optical path: n × t = 1.52 × 2.0 mm = 3.04 mm. The plate behaves like 1.04 mm of extra air path, which is the idea behind fringe-shift questions with a glass sheet in Young's experiment.
Worked example 3: from water into glass
The 633 nm light first enters water (n = 1.33), then goes from the water into crown glass.
In water: 633 / 1.33 = 476 nm
Water to glass: λ₂ = λ₁ × n₁/n₂ = 476 × 1.33/1.52 = 416 nm
That is the same 416 nm as going straight from air into glass. The wavelength depends only on the medium the light is in now, not on the route it took.
What mistakes do students make?
Changing the frequency. Frequency is fixed by the source. If your answer has a new frequency in glass, recheck.
Multiplying by n. Wavelength and speed both go down in a denser medium, so divide by n.
Thinking light needs energy to speed up again. When it leaves the glass it returns to c at once. Its speed is set by the medium.
Forgetting dispersion. NCERT notes that the refractive index also depends on wavelength, so violet light slows more than red in glass. One value of n is valid for one colour.
A practice routine
Write the three rules and the crest-counting reason from memory.
Repeat example 1 for 450 nm light in water. (Check: 338 nm, 2.25 × 10⁸ m/s, 6.66 × 10¹⁴ Hz.)
Find the thickness of glass (n = 1.52) that holds exactly 1000 more wavelengths of 633 nm light than the same thickness of air.
Do two refraction questions from our post on refraction at a spherical surface, then note which used n and which used λ.
If wavefronts slowing and bending is easier to see than read, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Start with practice step 3: it takes five minutes and tests all three rules at once.
