Question

What is the minimum (non-zero) thickness of a benzene (n = 1.501) thin film that will result in constructive interference when viewed at normal incidence and illuminated with orange light (λvacuum = 615 nm)? A glass slide (ng = 1.620) supports the thin film.

Answer 1

Explanation:

At each reflecting surface (benzene and glass) there will be 180 degree phase change.

Now, for constructive interference the optical path in benzene is $\lambda$.

Formula to calculate thickness of a benzene thin film is as follows.

     Optical path length through benzene ($\lambda$) = $2 \times n \times d$

Hence, substituting the given values into the above formula as follows.

    Optical path length through benzene = $2 \times n \times d$

                   d = $\frac{\text{Optical path length through benzene}}{2 \times n}$

                       = $\frac{\lambda}{2 \times n}$  

                       = $\frac{615 \times 10^{-9}}{2 \times 1.501}$   (as 1 nm = $10^{-9}m$

                       = 204.9 m          

Thus, we can conclude that minimum thickness of benzene is 204.9 m.