Blue light with a wavelength of 425 nm passes from a vacuum into a glass lens, and the index of refraction is found to be 1.65. The glass lens is replaced with a plastic lens. The index of refraction for the plastic lens is 1.54. In which one of the two lenses does the light have the greatest speed and what is that speed ?

Options

A. glass, `2.28xx10^(8)m//s`
B. plastic, `2.13xx10^(8)m//s`
C. glass, `1.82xx10^(8)m//s`
D. plastic, `1.95xx10^(8)m//s`

Answer

To solve the problem, we need to determine the speed of light in both the glass lens and the plastic lens, and then compare the two speeds to find out which one is greater.

### Step-by-Step Solution:

1. **Identify the known values:**
- Wavelength of blue light, \( \lambda = 425 \, \text{nm} \) (not directly needed for speed calculation).
- Speed of light in vacuum, \( c = 3 \times 10^8 \, \text{m/s} \).
- Refractive index of glass, \( n_g = 1.65 \).
- Refractive index of plastic, \( n_p = 1.54 \).

2. **Write the formula for the speed of light in a medium:**
The speed of light in a medium is given by the formula:
\[ v = \frac{c}{n} \]
where \( v \) is the speed of light in the medium, \( c \) is the speed of light in vacuum, and \( n \) is the refractive index of the medium.

3. **Calculate the speed of light in the glass lens:**
Using the refractive index of glass:
\[ v_g = \frac{c}{n_g} = \frac{3 \times 10^8 \, \text{m/s}}{1.65} \]
Performing the calculation:
\[ v_g \approx 1.818 \times 10^8 \, \text{m/s} \]

4. **Calculate the speed of light in the plastic lens:**
Using the refractive index of plastic:
\[ v_p = \frac{c}{n_p} = \frac{3 \times 10^8 \, \text{m/s}}{1.54} \]
Performing the calculation:
\[ v_p \approx 1.948 \times 10^8 \, \text{m/s} \]

5. **Compare the speeds:**
- Speed in glass, \( v_g \approx 1.818 \times 10^8 \, \text{m/s} \)
- Speed in plastic, \( v_p \approx 1.948 \times 10^8 \, \text{m/s} \)

Since \( v_p > v_g \), the light travels faster in the plastic lens.

6. **Final Answer:**
The light has the greatest speed in the plastic lens, and that speed is approximately:
\[ v_p \approx 1.948 \times 10^8 \, \text{m/s} \]