Wave Optics: Light in Water & Glass Wedge - Find Thickness/Angle

[lha08]

Homework Statement

Light of wavelength 640 nm in water illuminates a glass (n = 1.5) wedge submerged in water (n = 1.33). If the distance between successive bright fringes is 6mm.
Find a) the change in thickness of the glass between these fringes
and b) the angle of the wedge

Homework Equations

The Attempt at a Solution

Since it's a bright fringe and n=1 to n=1.5 to n=1.33, it's constructive so 2t= (m+1/2)lambda/nfilm. WHen i plug in the values i get t=(m+1/2)(6.40X10^-7 m)/2(2.5) but I'm not sure how i can find m with the distance between the bright fringes (y=6X10^-3 m)...
And for the angle of the wedge i thought that it would be pi since n=1 to n=1.5 to n=1.33.

 

[Doc Al]

Since it's a bright fringe and n=1 to n=1.5 to n=1.33,

Note that the wedge is submerged in water. Note also that the wavelength is given in water.

Between adjacent bright fringes, what must be the additional phase difference due to the thickness of the glass? What extra thickness is required to produce that phase difference?

 

[lha08]

Note that the wedge is submerged in water. Note also that the wavelength is given in water.

Between adjacent bright fringes, what must be the additional phase difference due to the thickness of the glass? What extra thickness is required to produce that phase difference?

so n=1.33 to n=1.5 to n=1.33 but my formula doesn't change? But i don't know how to use 6mm to get m...like the phase change formula is =2pi(path difference)/lambda or phase change=2pi(m)...? I'm kind of confused at that point..

 

[Doc Al]

it's constructive so 2t= (m+1/2)lambda/nfilm.

This formula is almost right. That formula applies when the wavelength is given in air. How would you modify it to use the given wavelength in water?

Then compare two adjacent fringes, call them:
t₁ at m = n
t₂ at m = n + 1

Find the change in thickness.