A beam of light of wavelength 633 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 constructive interference and maximum intensity, the path difference between the two beams of light coming from the two plates must be an integer multiple of the wavelength of the light.

In this case, the path difference is given by

2*d*cosθ,

where d is the air gap separation between the plates and θ is the angle between the incident beam and the normal to the plates. Since the incident beam is along the normal, θ = 0 and we get

path difference = 2*d.

For maximum intensity, this path difference must be an integer multiple of the wavelength, i.e.,

2*d = n*λ,

where n is an integer.

Substituting the values, we get

d = n*λ/2 = (1/2)*633 nm = 316.5 nm ≈ 317 nm.

Therefore, the transmitted beam will be of maximum intensity when the air gap separation between the plates is approximately 317 nm.

Hence, the correct choice is C.

A Beam of Light of Wavelength 633 Nm Is Incident