The wavelength of light in vacuum is 5000 `overset(o)(A)`. When it travels normally through diamond of thickness 1.0 mm find the number of waves of light in 1.0 mm of diamond. ( Refractive index of diamond = 2.417

Options

A. 4834 waves
B. 5834 waves
C. 4384 waves
D. 6834 waves

Explanation

To find the number of waves of light in 1.0 mm of diamond, we will follow these steps:

### Step 1: Convert the wavelength from angstroms to meters
The given wavelength of light in vacuum is 5000 angstroms. We need to convert this to meters for our calculations.

\[ 1 \text{ angstrom} = 10^{-10} \text{ meters} \]
Thus,
\[ \lambda = 5000 \text{ angstroms} = 5000 \times 10^{-10} \text{ meters} = 5 \times 10^{-7} \text{ meters} \]

### Step 2: Calculate the wavelength in diamond
When light travels through a medium, its wavelength changes according to the refractive index (μ) of that medium. The relationship is given by:

\[ \lambda' = \frac{\lambda}{\mu} \]

Given that the refractive index of diamond is 2.417, we can calculate the new wavelength (λ') in diamond:

\[ \lambda' = \frac{5 \times 10^{-7}}{2.417} \approx 2.07 \times 10^{-7} \text{ meters} \]

### Step 3: Convert the thickness of diamond to meters
The thickness of the diamond is given as 1.0 mm. We need to convert this to meters as well:

\[ T = 1.0 \text{ mm} = 1.0 \times 10^{-3} \text{ meters} \]

### Step 4: Calculate the number of waves in the diamond
The number of waves (N) that fit into the thickness of the diamond can be calculated using the formula:

\[ N = \frac{T}{\lambda'} \]

Substituting the values we have:

\[ N = \frac{1.0 \times 10^{-3}}{2.07 \times 10^{-7}} \approx 4831.4 \]

Since the number of waves must be a whole number, we round it to:

\[ N \approx 4831 \]

### Final Answer
The number of waves of light in 1.0 mm of diamond is approximately **4831**.

---