Question

if light travels from air (n=1), to a thin film of material M1 (n=1.45) on top of material M2 (n=1.72), determine the smallest thickness (in nm) of material M1 to observe constructive interference for the reflection of light of wavelength 619 nm

A. 13

B. 78

C. 213

D.129

E.386

2. an object is put in front of a diverging lens. The image is observed at a distance 3 cm from the lens. The height of the image is 4 cm. The magnitude of the focal length of the lens is 16.41 cm. The height of the object in cm is close to:

Answer 1

At the first interface, the incident light is in air, which has a lesser index of refraction than the thin film (material 1), so a phase shift occurs. At the second interface, the incident light is in the thin film (materiall), which has a lesser index of refraction than that of material 2, so a phase shift occurs. Because both reflections cause the same phase shift, they tend to put the waves in phase. The condition for constructive interference is, та t= 2 Where m=0,1,2,3,..... 2n The condition for destructive interference is, 1 2 t= m + Where m= 0,1,2,3,..... 22n Here 2 is the wavelength of light in vacuum, and nis the index of refraction of material 1.

The smallest thickness of the film (material 1) for reflection of light under constructive interference is, ma t= 2n From smallest thickness of the film we have m=1 . So the smallest thickness of the film (material 1) is, mi Ismallest 2n 1(619 nm) 2(1.45) = 213.448nm 213 nm