1. For a particular piece of glass the refractive index for X- rays of wavelength 1.60 X 10^-6 less than unity. At what maximum angle, measured to the surface , must a beam of X-rays strike the glass to undergo total internal reflection?
2 An oil drop of thickness 2.5 x 10^-5cm is on the surface of water in a water tank of area 1square meter. White light which is incident normally is observed through a spectrometer. The spectrum is seen to contain one dark band whose center has a wavelength of 6.5 x 10^-5cm in the air. Find the refractive index of oil.
To answer these questions, we will need to utilize the concepts of refraction and total internal reflection.
1. For the first question:
To determine the angle of incidence required for total internal reflection, we can use Snell's law and the definition of the critical angle. The critical angle is the angle of incidence at which the refracted ray is at an angle of 90 degrees to the surface, resulting in total internal reflection.
Let's assume the refractive index of the glass is n. According to the question, it is given that the refractive index for X-rays of wavelength 1.60 x 10^-6 is less than unity (less than 1).
Recall Snell's law:
n₁sinθ₁ = n₂sinθ₂
In this case, n₁ is the refractive index of the medium from which X-rays are coming (air or vacuum), and n₂ is the refractive index of the glass.
For total internal reflection to occur, the refracted angle (θ₂) would be 90 degrees. Thus, we can substitute sinθ₂ with 1 in Snell's law:
n₁sinθ₁ = n₂sin90
n₁sinθ₁ = n₂
Since it is mentioned that the refractive index for X-rays of wavelength 1.60 x 10^-6 is less than unity, we can write:
n₂ < 1
Now we can substitute the values and solve for the angle of incidence.
2. For the second question:
To find the refractive index of oil, we need to apply the concept of thin-film interference and the relationship between wavelength, refractive index, and thickness.
In this case, the oil drop forms a thin film on the surface of water. When light is incident on the thin film, it gets partially reflected from the air-oil and oil-water interfaces and also partially transmitted. These reflected and transmitted waves interfere, leading to the observation of dark or bright bands.
The formula for the dark band, which occurs when the path difference between the reflected waves is an odd multiple of half-wavelength, is given by:
2nt = (2m + 1)λ
Where:
n is the refractive index of the oil
t is the thickness of the oil drop
m is an integer representing the order of the dark band observed
λ is the wavelength of light in air
By substituting the given values (t = 2.5 x 10^-5 cm, m = 0, and λ = 6.5 x 10^-5 cm), we can solve the equation to find the refractive index of the oil.
Note: In both cases, it is important to ensure that the units are consistent for easier calculations.