physics---wave and light2

Since both R1 and R2 are reflected from a less dense to a dense medium, there is a phase change of pi radians (half a wavelength) on both interfaces. Hence, we have, for a bright fringe to occur,

2nt = m入
where n is the refractive index of the thin film
t is the tickness of the film
入 is the wavelength of light
m is a +ve integer

Let t1 and t2 be the thicknesses of the thin film on the left and right hand sides respectively,

Then, 2n(t1) = m入
and 2n(t2) = (m+15)入

Subtracting, 2n(t2 - t1) = 15入
t2 - t1 = 15 x 600/(2 x 1.25) nm = 3600 nm