Lesson 6 of 10Light & Geometric Optics

Unit 4 · Lesson 6 of 10

Snell's Law

The exact rule that links the angles on each side of a boundary to the indices of refraction, how to use it to find an angle or an unknown index, and what happens to a ray passing through a glass block.

3 min read8 practice questionsInteractiveAsk AI about this →

Key things to remember

1Snell's law: $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ , with both angles measured from the normal.
2Solve for an angle with $sin θ_{2} = \frac{n _{1} sin θ _{1}}{n _{2}}$ , then take $sin^{- 1}$ .
3From air: $n_{2} = \frac{sin θ _{1}}{sin θ _{2}}$ .
4 $\frac{n _{2}}{n _{1}} = \frac{sin θ _{1}}{sin θ _{2}} = \frac{v _{1}}{v _{2}}$
5A ray through a block with parallel faces exits parallel to its original direction, displaced sideways.

Knowing that light bends toward the normal in glass is useful, but a lens designer needs to know by exactly how much. In 984, the Persian scholar Ibn Sahl, working in Baghdad, described the rule in a treatise on burning mirrors and lenses. The Dutch mathematician Willebrord Snellius found it again in 1621, and it is now known by his name: Snell's law. It is one short equation, and with it you can predict the path of any ray across any boundary.

n_{1} sin θ_{1} = n_{2} sin θ_{2}

Snell's law: medium 1 is where the light comes from, medium 2 is where it goes. Both angles are measured from the normal.

Fig 1Snell's law in action: 45° in air (n = 1.00) becomes about 32° in water (n = 1.33), bending toward the normal; in the reverse direction, light from water into air bends away from the normal.

Read the equation as a balance: the product $n sin θ$ is the same on both sides of the boundary. If $n$ goes up, $sin θ$ must go down, so the angle gets smaller (toward the normal). If $n$ goes down, the angle gets larger (away from the normal). That is exactly the rule from the last lesson, now with numbers attached.

Draw a quick sketch with the normal and label medium 1 (incident side) and medium 2 (refracted side).
Write Snell's law and rearrange for the unknown before substituting, for example $sin θ_{2} = \frac{n _{1} sin θ _{1}}{n _{2}}$ .
Check that your calculator is in degree mode.
Calculate the sine value, then use inverse sine ( $sin^{- 1}$ ) to get the angle.
Check the direction: into a larger $n$ the angle must shrink; into a smaller $n$ it must grow.

Degree mode, and do not divide the angles

sin 4 5^{\circ}

comes out as

0.851

instead of

0.707

, your calculator is in radians. Also, you cannot cancel the 'sin' and write

θ_{2} = \frac{n _{1} θ _{1}}{n _{2}}

: the sine of an angle is not proportional to the angle, so you must take the sine first and the inverse sine last.

Other forms of the law

\frac{n _{2}}{n _{1}} = \frac{sin θ _{1}}{sin θ _{2}} = \frac{v _{1}}{v _{2}}

Because n = c/v, the ratio of the sines also equals the ratio of the speeds.

When medium 1 is air ( $n_{1} = 1.00$ ), Snell's law simplifies to $n_{2} = \frac{sin θ _{1}}{sin θ _{2}}$ . This is how the index of an unknown material is measured in the lab: shine a ray into a block, measure both angles with a protractor, and divide the sines. A graph of $sin θ_{1}$ against $sin θ_{2}$ is a straight line through the origin whose slope is the index.

A glass block shifts a ray sideways

When a ray passes through a rectangular block with parallel faces, it bends toward the normal on the way in and away from the normal by the same amount on the way out. The ray leaves travelling parallel to its original direction, but shifted sideways. This is called lateral displacement, and it is why a pencil seen through a thick window looks slightly offset.

Loading interactive…

Choose the two media and drag the angle of incidence; the Snell's law calculation and the refracted angle update together.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Light in air strikes a water surface (

n = 1.33

) at an angle of incidence of

45. 0^{\circ}

. Find the angle of refraction.

Example 2

A diver's flashlight beam travels up through water (

n = 1.33

) and meets the surface at

30. 0^{\circ}

to the normal. At what angle does it enter the air?

Example 3

In a lab, a ray in air strikes a transparent block at

50. 0^{\circ}

and refracts at

30. 0^{\circ}

inside it. Find the block's index of refraction and suggest what it is made of.

Example 4

A ray in air hits a rectangular glass block (

n = 1.50

) at

30. 0^{\circ}

. Find the angle inside the glass and the angle at which the ray leaves the far side.

Where marks get lost

Using radians mode, which makes every sine wrong. $sin 3 0^{\circ}$ must be exactly $0.5$ .
Cancelling the sines and treating the angles as proportional to $n$ . Take the sine first, do the arithmetic, then take the inverse sine.
Mixing up which medium is 1 and which is 2. Medium 1 is always the side the light is coming from.
Measuring the angles from the surface rather than the normal.
Not sanity-checking the direction: entering a larger index must give a smaller angle.

Flashcards 6

Snell's law

Snell's law solved for the refracted angle

Finding n of a material from air

Ratio form

+ 2 more

Practice

Light in air strikes glass (

n = 1.50

) at

30. 0^{\circ}

. Find the angle of refraction, to the nearest tenth of a degree.

Light in air enters crown glass (

n = 1.52

) at

60. 0^{\circ}

. Find the angle of refraction, to the nearest tenth of a degree.

A ray travels from glass (

n = 1.50

) into air, meeting the surface at

25. 0^{\circ}

. Find the angle of refraction in air, to the nearest tenth of a degree.

A ray in air at

35. 0^{\circ}

refracts to

25. 0^{\circ}

in a clear liquid. Find the index of refraction of the liquid, to 2 decimal places.

Light passes from water (

n = 1.33

) into crown glass (

n = 1.52

) at an angle of incidence of

40. 0^{\circ}

. Find the angle of refraction, to the nearest tenth of a degree.

A student computes

θ_{2} = \frac{n _{1} θ _{1}}{n _{2}}

instead of using sines. Why is this wrong?

A ray passes through a rectangular glass block with parallel faces. Compared with the ray going in, the ray coming out is:

Sort the steps for finding a refracted angle with Snell's law.

1Check that the ray bent the correct way
2Take the inverse sine to get $θ_{2}$
3Substitute and calculate the value of $sin θ_{2}$
4Identify medium 1 (where the light comes from) and medium 2
5Rearrange to $sin θ_{2} = \frac{n _{1} sin θ _{1}}{n _{2}}$

Spot a mistake on this page? Let us know