# Three problems concerning light interference and snells

1. Oct 10, 2011

### minerslave4

5. Two slits are separated by 0.195 mm. An interference pattern is formed on a screen 33.0 cm away by 656.3-nm light. Calculate the fraction of the maximum intensity a distance 0.600 cm away from the central maximum.

d =.000195m
L = .33 m
lambda = 656.3 nm
y = .006m

it's a small angle so sin = tan

by geometry, tan = y/L and sin = y/L

Intensity = Imax cos^2 (pi*d*sin/lambda) replace sin with y/L and numbers

I = Imax cos^2 (pi * .000195 * .006/.33 / 656.3e-9)
I get
I = .9146Imax

but it says my answer is off by a multiple of ten?

7. A thin film of oil (n = 1.30) is located on smooth, wet pavement. When viewed perpendicular to the pavement, the film reflects most strongly red light at 640 nm and reflects no yellow light at 548 nm. How thick is the oil film?

The layers are
Air n =1
oil n = 1.3
water n = 1.33
The path difference is equal to the thickness right

using 2tn = (m+.5)lambda for constructive interference and 2tn = m*lambda for destructive with constructive = 640nm, destructive = 548nm ?

dividing the equations by each other to find m:

constructive/destructive = (m+.5)/m
640/548 = 1 + 1/2m
m ~ 3

then plug back into one of equation to get t -> t = (3+.5)(640)/(2*1.30) = 861.54nm is this correct?

8. A material having an index of refraction of 1.25 is used as an antireflective coating on a piece of glass (n = 1.50). What should be the minimum thickness of this film in order to minimize reflection of 470 nm light?

this is destructive so

2tn = m*lambda

wait minimum reflection so m = 1? then if so

t = (1)(470nm)/(2)(1.25) = 188nm?

2. Oct 11, 2011

### SammyS

Staff Emeritus
Check Forum Rules about bumping. -- wait 24 hours. -- but you weren't terribly impatient.

7. Looks OK.