| << Chapter < Page | Chapter >> Page > |
Step 6. Solve the appropriate equation for the quantity to be determined (the unknown), and enter the knowns . Slits, gratings, and the Rayleigh limit involve equations.
Step 7. For thin film interference, you will have constructive interference for a total shift that is an integral number of wavelengths. You will have destructive interference for a total shift of a half-integral number of wavelengths . Always keep in mind that crest to crest is constructive whereas crest to trough is destructive.
Step 8. Check to see if the answer is reasonable: Does it make sense? Angles in interference patterns cannot be greater than , for example.
Which of the following best describes the cause of thin film interference?
(d)
A film of magnesium fluoride ( n = 1.38) is used to coat a glass camera lens ( n = 1.52). If the thickness of the film is 105 nm, calculate the wavelength of visible light that will have the most limited reflection.
What effect does increasing the wedge angle have on the spacing of interference fringes? If the wedge angle is too large, fringes are not observed. Why?
How is the difference in paths taken by two originally in-phase light waves related to whether they interfere constructively or destructively? How can this be affected by reflection? By refraction?
Is there a phase change in the light reflected from either surface of a contact lens floating on a person’s tear layer? The index of refraction of the lens is about 1.5, and its top surface is dry.
In placing a sample on a microscope slide, a glass cover is placed over a water drop on the glass slide. Light incident from above can reflect from the top and bottom of the glass cover and from the glass slide below the water drop. At which surfaces will there be a phase change in the reflected light?
Answer the above question if the fluid between the two pieces of crown glass is carbon disulfide.
While contemplating the food value of a slice of ham, you notice a rainbow of color reflected from its moist surface. Explain its origin.
An inventor notices that a soap bubble is dark at its thinnest and realizes that destructive interference is taking place for all wavelengths. How could she use this knowledge to make a non-reflective coating for lenses that is effective at all wavelengths? That is, what limits would there be on the index of refraction and thickness of the coating? How might this be impractical?
Notification Switch
Would you like to follow the 'College physics for ap® courses' conversation and receive update notifications?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|