Foundation5 past questions

Dispersion and rainbow formation

Study how light splits into its constituent colors through dispersion in prisms and natural phenomena like rainbows, including minimum deviation and dispersive power.

Why this shows up in the exam

NEET assesses your understanding of color separation and the physics behind natural optical displays.

How NEET tests this

Direct recall · 2 QsStatement analysis · 2 QsNumerical

Learn the idea

Dispersion separates white light into colours because refractive index varies with wavelength; in a water droplet the combination of refraction, internal reflection(s) and again refraction creates the rainbow arcs.

🧠 Memory hook: One bounce = primary (low), two hops = secondary (high) – the more hops, the higher the rainbow!

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Refractive index n(λ) decreases with increasing wavelength (red least, violet most)
  • Primary rainbow: one internal reflection, appears at ~42° from the antisolar point
  • Secondary rainbow: two internal reflections, appears at ~51° from the antisolar point and lies above the primary
  • Colour order is reversed in the secondary rainbow
  • Observer must have the Sun behind him to see a rainbow

How to approach it

  1. 1Read the statement and note how many internal reflections are mentioned
  2. 2Recall that one reflection → primary (lower arc), two reflections → secondary (higher arc)
  3. 3Use the angular positions (≈42° for primary, ≈51° for secondary) to decide which arc is above the other
  4. 4Check the colour‑order clue if given (reversed for secondary)

Worked example — watch it click

During a cloudy day, a primary and a secondary rainbow may be created, then the:

  • A)Primary rainbow is due to a double internal reflection and is formed above the secondary one.
  • B)Primary rainbow is due to a double internal reflection and is formed below the secondary one.
  • ✅Secondary rainbow is due to a double internal reflection and is formed above the primary one.
  • D)Secondary rainbow is due to a single internal reflection and is formed above the primary one.

The concept behind this problem

The question asks which rainbow results from double internal reflection and its relative position, directly testing the link between reflection count and angular height of the arcs.

Step by step

  1. 1Primary rainbow: formed by one internal reflection inside water droplets, appears at ~42° from antisolar point.
  2. 2Secondary rainbow: formed by two internal reflections, appears at ~51° from antisolar point (higher/above primary), with reversed color order and dimmer.
  3. 3Therefore, secondary rainbow is due to double internal reflection and is formed above the primary one.

Watch out

Students often answer that the secondary rainbow comes from a single internal reflection or that it lies below the primary.

Common slip-ups that cost marks

  • •Confusing the number of internal reflections for primary and secondary
  • •Assuming the secondary rainbow appears below the primary
  • •Ignoring the reversed colour order as a hint

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 5NEET 2019

Pick the wrong answer in the context with rainbow. (a) An observer can see a rainbow when his front is towards the sun. (b) Rainbow is a combined effect of dispersion refraction and reflection of sunlight. (c) When the light rays undergo two internal reflections in a water drop, a secondary rainbow is formed. (c) The order of colours is reversed in the secondary rainbow.

Push further

More challenging

8 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 8

An observer is viewing a rainbow. In which direction relative to the Sun must the observer be positioned to see a primary rainbow?