Consult Interactive Solution 27.18 to review a model for solving this problem. A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in visible light with wavelength 653 nm in vacuum. Assuming that the visible spectrum extends from 380 to 750 nm, what is the longest visible wavelength (in vacuum) for which the film will appear bright due to constructive interference

Explanation:

In the given case for destructive interference , the condition is,

path difference = (2n+1)λ /2  where n is an integer and λ is wavelength

2 μ d = (2n+1)λ /2

Putting λ = 653 nm

for minimum thickness n = 0

2 μ d = 653 / 2 nm

= 326.5 nm

For constructive interference the condition is

2 μ d = n λ₁

326.5 nm = n λ₁

λ₁ = 326.5 / n  

For n = 1

λ₁ = 326.5 nm ,

or , 326.5nm .

Longest wavelength possible is 326.5

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

increase masses, decrease separation