Final answer:
The correct option for the intensity of reflected light when a laser shines on an interface between vacuum and carbon dioxide with an index of refraction of 1.00045 is (b) 5.1 x 10^-8 W/m². This is calculated using Fresnel's equations and reflects the extremely low reflectance at normal incidence due to the close refractive indices.
Step-by-step explanation:
To calculate the intensity of the reflected light when a laser shines on an interface between vacuum and carbon dioxide, we need to use the Fresnel equations. However, given that carbon dioxide has a very close index of refraction to vacuum (1.00045), the reflection will be very low. The intensity of the reflected light can't typically be calculated directly without knowing the angle of incidence, but we can estimate it using the formula for reflectance at normal incidence (perpendicular to the surface):
R = ((n1 - n2) / (n1 + n2))^2
Where n1 is the refractive index of the first medium (vacuum, which is 1.00) and n2 is the refractive index of carbon dioxide (1.00045).
This calculation gives us:
R = ((1.00 - 1.00045) / (1.00 + 1.00045))^2 = ((-0.00045) / (2.00045))^2 ≈ 5.1 x 10^-8
Then, the intensity of the reflected light (Ir) is:
Ir = R * Ii
Where Ii is the intensity of the incident light (1 W/m²).
So the reflected intensity is:
Ir = 5.1 x 10^-8 * 1 W/m² = 5.1 x 10^-8 W/m²
The correct option from the given choices is (b) 5.1 x 10^-8 W/m².