Question

A uniform film of TiO2, 1036 nm thick and having index of refraction 2.62, is spread uniformly over the surface of crown glass of refractive index 1.52. Light of wavelength 545 nm falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels.

Part A
What is the minimum thickness of TiO2 that you must add so the reflected light cancels as desired?
Part B
After you make the adjusment in part (a), what is the path difference between the light reflected off the top of the firm and the light that cancels it after traveling through the film? Express your answer in nanometers.
Part C
Express your answet in wavelengths of the light in the TiO2 film.

Answer 1

To cancel the reflection of light, the minimum thickness of the TiO2 film needs to be calculated using the equation 2t = (1/2)λ / n.

Step-by-step explanation:

In order to cancel the reflection of light, we need to create destructive interference. The condition for destructive interference is that the path difference between the two reflected rays is half a wavelength. The path difference is equal to 2t, where 't' is the thickness of the film.

Given that the wavelength of light is 545 nm and the refractive indices of the film and glass are 2.62 and 1.52 respectively, we can use the equation:

2t = (m + 1/2)λ / n

Where 'm' is an integer representing the number of wavelengths the light travels inside the film.

For destructive interference, we want the path difference to be 1/2 wavelength, so we can set 'm' to zero:

2t = (1/2)λ / n

Plugging in the values, we can solve for 't' to find the minimum thickness needed to cancel the reflection.

Answer 2

To cancel the reflected light, the minimum thickness of TiO2 that must be added is determined by the path difference between the reflected light and the light that travels through the film. The path difference is equal to half the wavelength of the light.

Step-by-step explanation:

To cancel the reflected light, we need to create destructive interference between the light reflected from the top surface of the film and the light that passes through the film and reflects from the bottom surface. For destructive interference to occur, the path difference between the two reflected rays must be equal to half the wavelength of the light.

Using the formula for the path difference, which is 2nt cos(theta), where n is the index of refraction of the film, t is the thickness of the film, and theta is the angle of incidence (which is 0 for normal incidence), we can solve for the minimum thickness of the film.

For part A of the question, we have n1 = 1 (air), n2 = 2.62 (TiO2), n3 = 1.52 (crown glass). Plugging these values into the formula, we get 2(2.62)(t) = 0.5(545 nm). Solving for t gives us the minimum thickness of TiO2 that needs to be added.