Calculating Angle of Wedge for Fringe Separation

[fredrick08]

Homework Statement

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?

Homework Equations

t(x)=xtan(theta)
2nt=m*lamda (m+.5) for deconstructive
x=(lamda/2)/2n*tan(theta)
t=lamda/2n

The Attempt at a Solution

ok I am 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?

 

[fredrick08]

anyone?

 

[fredrick08]

anyone know how?

 

[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

 

[fredrick08]

ok thankyou

 

[fredrick08]

hi again... srry, I've been looking at wat u said, andi don't understand how t=x(theta) isn't it xtan(theta)?? and from then i thought tan(theta)=t/x?? but it doesn't work because i have 2 unknowns? theta and x. t=lamda/2n... please can u tell me the process to solve this problem?

 

[fredrick08]

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

 

[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 wavelength of extra 't'