## Snell’s Law

Consider sunlight travelling through the air strikes a flat slab of glass at an incident angle of 25°. If the glass’s index of refraction is 2, what is the light’s angle of refraction at the moment it leaves the glass? Note the index of refraction for air is 1.

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__**Hint**

**Hint**

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__**Hint 2**

**Hint 2**

Snell’s Law:

$$$n_isin\theta_i=n_rsin\theta_r$$$

where
$$n_i$$
is the incident index,
$$n_r$$
is the refracted index,
$$\theta_i$$
is the incident angle, and
$$\theta_r$$
is the refracted angle.

Snell’s Law:

$$$n_isin\theta_i=n_rsin\theta_r$$$

where
$$n_i$$
is the incident index,
$$n_r$$
is the refracted index,
$$\theta_i$$
is the incident angle, and
$$\theta_r$$
is the refracted angle.

First, let’s find the refracted angle inside the glass slab:

$$$(1)sin25^{\circ}=(2)sin\theta_r$$$

$$$sin\theta_r=\frac{0.4226}{2}=0.211$$$

$$$\theta_r=sin^{-1}(0.211)=12.2^{\circ}$$$

Now that we determined the refracted angle inside the glass lab, let’s reuse Snell’s law again to determine
$$\theta'$$
:

$$$(2)sin12.2^{\circ}=(1)sin\theta'_r$$$

$$$sin\theta'_r=\frac{(2)sin12.2^{\circ}}{1}=0.4226$$$

$$$\theta'_r=sin^{-1}(0.4226)=25^{\circ}$$$

The sunlight exits the glass at an angle of refraction of 25°, which is the same angle it entered.

25°