Light bends towards normal entering denser medium. Calculate angle of refraction.

Snell's Law of Refraction — n₁ sin i = n₂ sin r

Question

A ray of light travels from air (refractive index 1.0) into glass (refractive index 1.5). The angle of incidence is 30°. Find the angle of refraction.

Solution — Step by Step

We have two media: air and glass. The light is going from air to glass — this means it’s going from a rarer medium to a denser medium, so the ray will bend towards the normal.

$n_1 = 1.0$ (air)
$n_2 = 1.5$ (glass)
$i = 30°$ , find $r = ?$

Snell’s Law relates the two angles through the refractive indices:

n_1 \sin i = n_2 \sin r

This equation comes directly from the fact that light slows down when entering a denser medium, and the wavefronts must remain connected at the boundary — the frequency doesn’t change, but the wavelength does.

1.0 \times \sin 30° = 1.5 \times \sin r

We know $\sin 30° = 0.5$ , so:

1.0 \times 0.5 = 1.5 \times \sin r

\sin r = \frac{0.5}{1.5} = \frac{1}{3} \approx 0.333

r = \sin^{-1}(0.333) \approx 19.47°

We round this to approximately 19.5° in board exams. Some books write $\approx 20°$ — either is acceptable.

Why This Works

Snell’s Law is really a statement about how light conserves momentum parallel to the boundary surface. When light slows down in glass, the component of velocity along the surface must remain the same — and this forces the angle to change.

The ratio $n_2 / n_1$ tells us exactly how much the angle compresses. Here, $1.5 / 1.0 = 1.5$ , meaning $\sin r$ is 1.5 times smaller than $\sin i$ . Smaller $\sin r$ means smaller $r$ , which confirms the ray bends towards the normal.

Quick sanity check: light going into a denser medium ( $n_2 > n_1$ ) always gives $r < i$ . If your answer shows $r > i$ for this scenario, something went wrong in the calculation.

For the CBSE Class 10 pattern, you’re almost always given a 30°, 45°, or 60° angle of incidence — all of which have clean $\sin$ values. This question type has appeared in board papers multiple times since 2018.

Alternative Method — Using the Refractive Index Ratio Directly

Instead of plugging into the formula immediately, we can rearrange first:

\sin r = \frac{n_1}{n_2} \sin i = \frac{1.0}{1.5} \times \sin 30° = \frac{2}{3} \times 0.5 = \frac{1}{3}

Writing $n_1/n_2$ as a fraction first ( $2/3$ ) keeps the numbers cleaner than working with decimals. This is especially helpful in JEE Main where you don’t have a calculator and messy decimals slow you down.

r = \sin^{-1}\!\left(\frac{1}{3}\right) \approx 19.47°

Same answer, cleaner arithmetic path.

Common Mistake

Swapping $n_1$ and $n_2$ — students write $n_2 \sin i = n_1 \sin r$ and get $\sin r = 0.75$ , giving $r \approx 48.6°$ . This is larger than the angle of incidence, which contradicts the rule that light bends towards normal in a denser medium. Always pair $n_1$ with $\sin i$ (the incident side) and $n_2$ with $\sin r$ (the refracted side).

A quick check: if your $r > i$ when going from rarer to denser, flip your $n$ values.

n_1 \sin i = n_2 \sin r

Where:

$n_1$ = refractive index of medium where the incident ray travels
$n_2$ = refractive index of medium where the refracted ray travels
$i$ = angle of incidence (measured from normal)
$r$ = angle of refraction (measured from normal)

Final answer: $r \approx 19.47°$ (≈ 19.5° for board exams)

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Refractive Index Ratio Directly

Common Mistake

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