# Optics qn

1. Mar 26, 2009

### fredrick08

1. The problem statement, all variables and given/known data
fringes are observed when a parrallel beam of light of wavelength 500nm is incident perpendiculary onto a wedge-shaped film with an index of refraction of 1.5. What is the angle of the wedge if the fringe separation is 1/3cm?

2. Relevant equations
t(x)=xtan(theta)
2nt=m*lamda (m+.5) for deconstructive
x=(lamda/2)/2n*tan(theta)
t=lamda/2n
3. The attempt at a solution
ok im really struggling to understand this qn, i thought simple trig, t=1.66x10^-9
but now completely stuck... coz i think i need the x value..... but no idea how to find it??? can anyone help?

2. Mar 26, 2009

### fredrick08

anyone?

3. Mar 26, 2009

### fredrick08

anyone know how?

4. Mar 26, 2009

### mgb_phys

Sorry - if you reply to your own question is shows as answered so people ignore it!

With$$\theta$$ in radians t = x $$\theta$$
As you say you need 2t to be an odd number of half wavelenghts
t = n(m+0.5) $$\lambda$$

Now you have two values of x for m and m+1

5. Mar 26, 2009

### fredrick08

ok thankyou

6. Mar 26, 2009

### fredrick08

hi again... srry, ive been looking at wat u said, andi dont understand how t=x(theta) isnt it xtan(theta)?? and from then i thought tan(theta)=t/x?? but it doesnt work because i have 2 unknowns??? theta and x. t=lamda/2n... plz can u tell me the process to solve this problem?

7. Mar 26, 2009

### fredrick08

but then where does this d=1/3cm come from???

8. Mar 26, 2009

### mgb_phys

For small angles, angle = sin angle = tan angle (assuming radians)
d (in your diagram) doens't equal 1/3cm, that is the spacing of the fringes along the wedge - essentially it's the extra length of wedge needed to get one wavelenght of extra 't'