Urgent help please, physics test tomorrow (Light/waves)?

Question

A air wedge between two microscope slides 10cm long and separated at one end by a paper of thickness 0.08mm is illuminated with red light of wavelength 660nm.

a. What is the spacing of the dark fringes in the interference pattern reflected from the air wedge?

b. How would the spacing change if the wedge were filled with water (n=1.33)?

Answer 1

There will be dark fringes when the reflected light travels 330nm + p*660nm further from the top of the bottom slide than the bottom of the top slide, where p is an integer.

The light  travels the distance between the two slides twice, once down and again back up, so the distance between the slides is (330nm + p*660nm)/2.

The distance between the slides, d(x) vs the distance where they touch is, d(x) = (0.08/100)x.

The first dark fringe happens when p=0; (330nm + 0*660nm)/2 = (0.08/100)x.

x = 330*100/0.08 = 412500nm = 0.4125 mm.

The second dark fringe happens when p=1; (330nm + 1*660nm)/2 = (0.08/100)x.

x = 990*100/0.08 = 1237500nm = 1.237500 mm.

1.237500 mm - 0.4125 mm = 0.825000 mm.

a. The spacing between the fringes is 0.825000 mm.

b. If water were filled the space between the slides, the spacing would be 1.33x0.825000 mm = 1.09725 cm.

Answers should be rounded off to

a. 0.8mm

b. 1.1cm