## Angle of Refraction

A child underwater in a pool shines a flashlight towards the surface at a 40° angle from the vertical. What angle does the beam of light leave the pool? Note the index of refraction for water and air are 1.33 and 1.00 respectively.

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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.

Thus, the refracted angle is:

$$$(1.33)sin40^{\circ}=(1.00)sin\theta_r$$$

$$$sin\theta_r=\frac{0.855}{1.00}=0.855$$$

$$$\theta_r=sin^{-1}(0.855)=58.7^{\circ}$$$

58.7°