To determine which wavelengths will be amplified in the reflected beam, we need to consider the phenomenon of thin-film interference. When light passes through a thin film, such as a glass plate, and reflects from its boundaries, constructive or destructive interference can occur, depending on the wavelength of the incident light.
In this case, the glass plate has a thickness of 0.5 µm, and the refractive index of the glass is 1.5. When light passes from one medium (air) to another (glass), it undergoes both reflection and transmission. Some light is reflected at the air-glass interface, while the rest is transmitted into the glass.
For constructive interference to occur, the path length difference between the reflected rays from the front and back surfaces of the glass plate must be an integral multiple of the wavelength. Mathematically, this condition can be expressed as:
2t * n = m * λ
where: t is the thickness of the glass plate, n is the refractive index of the glass, m is an integer representing the order of interference, λ is the wavelength of light.
Since the incident light is white light, which consists of a range of wavelengths, different wavelengths will interfere constructively at different angles.
To find the wavelengths that will be amplified in the reflected beam, we can consider the visible spectrum, which ranges approximately from 400 nm (violet) to 700 nm (red). We can calculate the wavelengths that satisfy the constructive interference condition for the given glass plate thickness and refractive index.
Let's substitute the given values into the equation and solve for λ:
2 * 0.5 µm * 1.5 = m * λ
1.5 µm = m * λ
λ = (1.5 µm) / m
Now, we can calculate the wavelengths for different orders of interference (m) within the visible spectrum:
For m = 1: λ = (1.5 µm) / 1 = 1.5 µm (This falls within the infrared range and is not visible.)
For m = 2: λ = (1.5 µm) / 2 = 0.75 µm (This falls within the infrared range and is not visible.)
For m = 3: λ = (1.5 µm) / 3 = 0.5 µm (This falls within the visible range, corresponding to green light.)
For m = 4: λ = (1.5 µm) / 4 = 0.375 µm (This falls within the ultraviolet range and is not visible.)
From this calculation, we find that the only wavelength within the visible part of the spectrum that will be amplified in the reflected beam is approximately 0.5 µm, corresponding to green light.
Therefore, when a beam of white light falls perpendicularly on a glass plate with a thickness of 0.5 µm and a refractive index of 1.5, only the green wavelength will be amplified in the reflected beam due to constructive interference.