Answer to Question #102423 in Optics for ROHIT SHARMA

Question

A wedge shaped film has refractive index 1.34 and thickness of extreme sides 0 (zero) and t. If a light of wavelength 492 nm is incident normally on it and 20 fringes are obtained, determine t.

Answer 1

$2t\sqrt{n^2-\sin^2i}-\frac{\lambda}{2}=m\lambda$

$2tn-\frac{\lambda}{2}=m\lambda$

$t=\frac{(m\lambda+ \frac{\lambda}{2})}{2n}=\frac{(20\cdot 492\cdot 10^{-9}+ \frac{492\cdot 10^{-9}}{2})}{2\cdot 1.34}=3.76\cdot 10^{-6} m=3.76\mu m$