Snell's Law examples and solutions

Hello there! 👋 It's fantastic you're diving into Snell's Law – it's a fundamental concept in optics that helps us understand so much about how light behaves. Don't worry, many find it tricky at first, but with a few clear examples, it really clicks! Let's break it down. 😊

What is Snell's Law?

At its core, Snell's Law describes the relationship between the angles of incidence and refraction for a light ray passing through the boundary between two different isotropic media, such as air and water. When light travels from one medium to another, it often changes direction – this phenomenon is called refraction. Snell's Law quantifies this bending! 💡

The formula is expressed as:

$n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$

Where:

• $n_1$ is the refractive index of the first medium (where light originates).
• $\theta_1$ is the angle of incidence (the angle between the incoming light ray and the normal to the surface).
• $n_2$ is the refractive index of the second medium (where light enters).
• $\theta_2$ is the angle of refraction (the angle between the refracted light ray and the normal to the surface).

The normal is an imaginary line perpendicular to the surface at the point where the light ray strikes it. Remember, angles are always measured with respect to the normal! 📐

Example 1: Light Entering Water from Air

Imagine a light beam shining from air into a calm swimming pool. How much does it bend?

Given:

• Refractive index of air ($n_{\text{air}}$) $\approx 1.00$
• Refractive index of water ($n_{\text{water}}$) $\approx 1.33$
• Angle of incidence ($\theta_1$) $= 30^{\circ}$

Solution:

Using Snell's Law: $n_{\text{air}} \sin(\theta_1) = n_{\text{water}} \sin(\theta_2)$

Substitute the known values:

$1.00 \cdot \sin(30^{\circ}) = 1.33 \cdot \sin(\theta_2)$

Calculate $\sin(30^{\circ})$, which is $0.5$:

$1.00 \cdot 0.5 = 1.33 \cdot \sin(\theta_2)$
$0.5 = 1.33 \cdot \sin(\theta_2)$

Now, solve for $\sin(\theta_2)$:

$\sin(\theta_2) = \frac{0.5}{1.33} \approx 0.3759$

Finally, find $\theta_2$ by taking the inverse sine (arcsin):

$\theta_2 = \arcsin(0.3759) \approx 22.09^{\circ}$

Result: The light ray bends towards the normal, from $30^{\circ}$ to approximately $22.09^{\circ}$. This makes sense because light slows down when going from a less optically dense medium (air) to a more optically dense medium (water).

Example 2: Light Exiting Glass into Air

Consider light inside a glass prism exiting into the surrounding air.

Given:

• Refractive index of glass ($n_{\text{glass}}$) $\approx 1.50$
• Refractive index of air ($n_{\text{air}}$) $\approx 1.00$
• Angle of incidence inside the glass ($\theta_1$) $= 35^{\circ}$

Solution:

Using Snell's Law: $n_{\text{glass}} \sin(\theta_1) = n_{\text{air}} \sin(\theta_2)$

Substitute the known values:

$1.50 \cdot \sin(35^{\circ}) = 1.00 \cdot \sin(\theta_2)$

Calculate $\sin(35^{\circ})$, which is approximately $0.5736$:

$1.50 \cdot 0.5736 = 1.00 \cdot \sin(\theta_2)$
$0.8604 = \sin(\theta_2)$

Solve for $\theta_2$:

$\theta_2 = \arcsin(0.8604) \approx 59.36^{\circ}$

Result: The light ray bends away from the normal, from $35^{\circ}$ to approximately $59.36^{\circ}$. This happens because light speeds up when going from a more optically dense medium (glass) to a less optically dense medium (air). If $\theta_1$ were too large here, you might even get Total Internal Reflection! ✨

I hope these examples help clarify Snell's Law for you! Keep practicing, and you'll master it in no time! Let me know if anything is unclear. 😊