Answer:
t = 103.45 n m
Explanation:
given,
refractive index of cornea = 1.38
refractive index of eye drop = 1.45
wavelength of refractive index = 600 nm
refractive index of eye drop is greater than refractive index of cornea and the air.
Formula used in this case
for constructive interference
[tex]2 n t = (m + \dfrac{1}{2})\lambda[/tex]
At m = 0 for the minimum thickness, so
[tex]2\times 1.45 \times t = (0 + 0.5)\times 600[/tex]
[tex]2.9 \times t =300[/tex]
[tex] t =\dfrac{300}{2.9}[/tex]
t = 103.45 n m
the minimum thickness of the film of eyedrops t = 103.45 n m