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 520.0 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. What is the minimum thickness of TiO2 that you must add so the reflected light cancels as desired?

Accepted Answer

Step 1: Understand the concept of thin-film interference. Reflected light cancels when destructive interference occurs. This happens when the optical path difference between the two reflected rays is an odd multiple of half the wavelength in the film. Step 2: Calculate the wavelength of light inside the TiO2 film. The wavelength in the film is given by $$ \lambda_{film} = \frac{\lambda_{air}}{n_{film}} $$, where $$ \lambda_{air}  is $$the wavelength in air (520.0 nm) and $$ n_{film}  is $$the refractive index of TiO2 (2.62). Step 3: Determine the condition for destructive interference. For destructive interference, the optical path difference must satisfy $$ 2t = (m + \frac{1}{2}) \lambda_{film} $$, where $$ t  is $$the thickness of the film, $$ m  is an $$integer (order of interference), and $$ \lambda_{film}  is $$the wavelength in the film. Step 4: Calculate the additional thickness needed. The current thickness of the film is 1036 nm. To achieve destructive interference, find the minimum additional thickness $$ \Delta t $$ such that the total thickness $$ t_{total} = t_{current} + \Delta t $$ satisfies the interference condition. Step 5: Solve for $$ \Delta t . $$Rearrange the interference equation to isolate $$ \Delta t $$, ensuring that the total thickness corresponds to the next odd multiple of half the wavelength in the film. Use $$ \Delta t = \frac{(m + \frac{1}{2}) \lambda_{film}}{2} - t_{current}  to $$find the minimum additional thickness.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Interference of Light

Refractive Index

Thin Film Thickness