Question

A lens appears greenish yellow (λ=570nm is strongest) when white light reflects from it. What minimum thickness of coating (n=1.30) do you think is used on such a glass (n=1.51) lens?

Answer 1

The minimum thickness of the coating on the lens is approximately 356.25 nm.

Step-by-step explanation:

To determine the minimum thickness of the coating, we can use the formula for destructive interference:

2t = λ / (2n - 1)

Where t is the thickness of the coating, λ is the wavelength of the light, and n is the index of refraction of the lens.

Given that the wavelength of the strongest light reflected is 570 nm, we can substitute the values into the formula:

2t = 570 nm / (2 * 1.3 - 1)

Simplifying, we get:

2t ≈ 570 nm / 1.6

t ≈ 356.25 nm

Therefore, the minimum thickness of the coating is approximately 356.25 nm.

Answer 2

The minimum thickness of the coating for the lens is approximately 1.018 ×
`10^(^-^7)` meters.

OPD = m * λ / 2

where:

m = an integer (1, 2, 3, ...)

λ = the reflected wavelength

OPD = 2t - 2r

λ = 570 nm, we want the smallest possible OPD that satisfies the condition above occurs when m = 1:

2t - 2r = λ / 2

= 570 nm / 2

= 285 nm

r = t * (n_coating / n_glass)

= t * (1.30 / 1.51)

= 0.86 t

Substituting this into the OPD equation:

2t - 2 * 0.86t = 285 nm

t = 1.018 *
`10^(^-^7^)` m