Blue light of wavelength 480 nm is most strongly reflected from a thin...
Assuming the boundaries are vacuum-oil, oil-glass, then there will be a phase shift at the vacuum oil boundary since the refractive index of vacuum n
vac < n
oil .. & another phase shift at the oil glass boundary because n
oil n
glass That tells us that constructive interference happens for integer wavelength mλ.
Light crosses a distance d in
but since its speed is lower in oil,
Then
∴ The condition for constructive interference becomes 2 n
oild = mλ
The next minimum thickness other than zero is when m = -1
So,
The correct answer is: 200
View all questions of this test
Blue light of wavelength 480 nm is most strongly reflected from a thin...
To determine the conditions for maximum reflection, we can use the equation for the optical path difference in a thin film:
2nt = mλ,
where n is the index of refraction of the medium the light is incident from (in this case, air), t is the thickness of the film, m is an integer representing the order of the reflection (m = 0 for the first order), and λ is the wavelength of light.
In this case, since the light is incident from air, the index of refraction is 1.
For maximum reflection of blue light with a wavelength of 480 nm, we can set up the equation as:
2(1)(t) = m(480 nm).
Since the question states that the blue light is most strongly reflected, we can assume that this corresponds to the first order reflection (m = 1).
Therefore, we have:
2t = 480 nm.
Simplifying, we find:
t = 240 nm.
So, the thickness of the thin film of oil on the glass slide that results in maximum reflection of blue light is 240 nm.