A beam of light of wavelength 674 nm is incident along the normal to two closely spaced parallel glass plates.For what air gap separation between the plates will the transmitted beam be of maximum intensity? (1 nm = 10−⁹ m)

Question

Quizplus · Accepted Answer

For maximum intensity of the transmitted beam, the path difference between the two waves must be equal to an integer multiple of the wavelength.
Path Difference = 2t cosθ = mλ, where t is the thickness of the glass plates, θ is the angle of incidence and m is an integer.
For normal incidence, cosθ = 1, so Path Difference = 2t = mλ
For maximum intensity, m = 0 (since it is not possible for t to be negative)
Therefore, 2t = 0, which implies t = 0. Hence, the air gap separation between the plates should be zero.
But in reality, it is not possible for the plates to touch each other. So, the maximum intensity will occur when the plates are as close as possible, i.e. when the thickness of the air gap (t) is equal to half the wavelength.
Therefore, t = λ/2 = 337 nm. (since the given wavelength is 674 nm)

A Beam of Light of Wavelength 674 Nm Is Incident