Part 1: Refractive Index of Glass

We can use Snell's Law to solve for the refractive index of the glass. Snell's Law states:

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

Where:

• $n_1$ is the refractive index of air (which is approximately 1),
• $\theta_1$ is the angle of incidence in air,
• $n_2$ is the refractive index of glass (the value we want to find),
• $\theta_2$ is the angle of refraction in glass.

Given data:

• $n_1 = 1$ (since it's air),
• $\theta_1 = 50^\circ$ (angle of incidence in air),
• $\theta_2 = 30^\circ$ (angle of refraction in glass).

Applying Snell's Law:

$1 \times \sin(50^\circ) = n_2 \times \sin(30^\circ)$

Now, solve for $n_2$:

$n_2 = \frac{\sin(50^\circ)}{\sin(30^\circ)}$

We know that $\sin(50^\circ) \approx 0.7660$ and $\sin(30^\circ) = 0.5$.

$n_2 = \frac{0.7660}{0.5} = 1.532$

So, the refractive index of the glass is approximately 1.532.

Part 2: Angle of Refraction for a 60° Angle of Incidence

Now, we need to find the angle of refraction when the angle of incidence is changed to 60°.

We use Snell's Law again:

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

Substitute the known values:

• $n_1 = 1$ (refractive index of air),
• $n_2 = 1.532$ (refractive index of glass),
• $\theta_1 = 60^\circ$ (new angle of incidence),
• $\theta_2$ is the angle of refraction in glass, which we need to find.

Using Snell's Law:

$\sin(60^\circ) = 1.532 \times \sin(\theta_2)$

$\sin(60^\circ) \approx 0.8660$

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

$\sin(\theta_2) = \frac{\sin(60^\circ)}{1.532} = \frac{0.8660}{1.532} \approx 0.565$

Now find $\theta_2$ by taking the inverse sine:

$\theta_2 = \sin^{-1}(0.565) \approx 34.6^\circ$

Thus, the angle of refraction is approximately 34.6°.

Final Answer:

• The refractive index of the glass is 1.532.
• The angle of refraction when the angle of incidence is 60° is approximately 34.6°.

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Would you like more details or clarifications on any part? Here are 5 related questions for further exploration:

1. How would the refractive index change if the angle of incidence were larger than the critical angle?
2. How does the refractive index affect the speed of light through a medium?
3. What is the effect of a larger refractive index on the bending of light?
4. How would the angle of refraction change if the material was denser than glass?
5. How can we calculate the critical angle for total internal reflection?

Tip: When dealing with Snell's law, always ensure the angles are in the correct units (degrees or radians), and remember that the refractive index of air is nearly always taken as 1 unless otherwise stated.