A light ray falling at an angle of` 45^@` with the surface of a clean slab of ice of thickness 1.00 m is refracted into it at an angle of 30°. Calculate the time taken by the light rays to cross the slab. Speed of light in vacuum `= 3 xx 10^8ms^-1`

Explanation

To solve the problem of calculating the time taken by a light ray to cross a slab of ice, we will follow these steps:

### Step 1: Understand the given information
- The angle of incidence (i) = 45°
- The angle of refraction (r) = 30°
- Thickness of the ice slab (t) = 1.00 m
- Speed of light in vacuum (c) = \(3 \times 10^8 \, \text{ms}^{-1}\)

### Step 2: Use Snell's Law to find the refractive index of ice
Snell's Law states:
\[ n_1 \sin(i) = n_2 \sin(r) \]
Where:
- \(n_1\) = refractive index of air = 1
- \(n_2\) = refractive index of ice
- \(i = 45°\)
- \(r = 30°\)

Substituting the known values:
\[ 1 \cdot \sin(45°) = n_2 \cdot \sin(30°) \]
\[ \sin(45°) = \frac{1}{\sqrt{2}}, \quad \sin(30°) = \frac{1}{2} \]
Thus, we have:
\[ \frac{1}{\sqrt{2}} = n_2 \cdot \frac{1}{2} \]
Rearranging gives:
\[ n_2 = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 3: Calculate the speed of light in ice
The speed of light in a medium is given by:
\[ v = \frac{c}{n} \]
Where:
- \(c = 3 \times 10^8 \, \text{ms}^{-1}\)
- \(n = \sqrt{2}\)

Substituting the values:
\[ v = \frac{3 \times 10^8}{\sqrt{2}} \, \text{ms}^{-1} \]

### Step 4: Calculate the time taken to cross the slab
The time taken (T) to cross the slab can be calculated using:
\[ T = \frac{t}{v} \]
Where:
- \(t = 1.00 \, \text{m}\)
- \(v = \frac{3 \times 10^8}{\sqrt{2}} \, \text{ms}^{-1}\)

Substituting the values:
\[ T = \frac{1.00}{\frac{3 \times 10^8}{\sqrt{2}}} \]
This simplifies to:
\[ T = \frac{1.00 \cdot \sqrt{2}}{3 \times 10^8} \]
\[ T = \frac{\sqrt{2}}{3 \times 10^8} \, \text{s} \]

### Step 5: Final calculation
Now we can calculate the numerical value:
\[ \sqrt{2} \approx 1.414 \]
Thus:
\[ T \approx \frac{1.414}{3 \times 10^8} \approx 4.71 \times 10^{-9} \, \text{s} \]

### Final Answer:
The time taken by the light rays to cross the slab is approximately \(4.71 \times 10^{-9} \, \text{s}\).

---