If the speed of light in ice is `2.3xx10^8 m//s,` what is its index of refraction? What is the critical angle of incidence for light going from ice to air?

Explanation

To solve the problem step by step, we will first find the index of refraction of ice and then calculate the critical angle of incidence for light transitioning from ice to air.

### Step 1: Identify the given values
- Speed of light in ice (V) = \(2.3 \times 10^8 \, \text{m/s}\)
- Speed of light in vacuum (C) = \(3.0 \times 10^8 \, \text{m/s}\)

### Step 2: Calculate the index of refraction (μ)
The index of refraction (μ) is given by the formula:
\[ \mu = \frac{C}{V} \]
Substituting the values:
\[ \mu = \frac{3.0 \times 10^8 \, \text{m/s}}{2.3 \times 10^8 \, \text{m/s}} \]

### Step 3: Perform the calculation
Calculating the above expression:
\[ \mu = \frac{3.0}{2.3} \approx 1.3043 \]
Rounding to two decimal places, we find:
\[ \mu \approx 1.30 \]

### Step 4: Calculate the critical angle (θc)
The critical angle (θc) can be calculated using the formula:
\[ \theta_c = \sin^{-1}\left(\frac{1}{\mu}\right) \]
Substituting the value of μ:
\[ \theta_c = \sin^{-1}\left(\frac{1}{1.30}\right) \]

### Step 5: Perform the calculation for the critical angle
Calculating the value:
\[ \frac{1}{1.30} \approx 0.7692 \]
Now, finding the inverse sine:
\[ \theta_c = \sin^{-1}(0.7692) \]
Using a calculator, we find:
\[ \theta_c \approx 50.1^\circ \]

### Final Answers:
- The index of refraction of ice (μ) is approximately **1.30**.
- The critical angle of incidence for light going from ice to air (θc) is approximately **50.1°**.