Question

What is the minimum index of refraction of a clear material if a minimum thickness of 121 nm , when laid on glass, is needed to reduce reflection to nearly zero when light of 675 nm is incident normally upon it? Assume that the film has an index less than that of the glass.

Answer 1

Answer:

$\mu = 1.39$

Explanation:

Since the reflection is nearly zero intensity

so here we will say that the reflected light must show destructive interference

so here we have

$path \: difference = \frac{\lambda}{2}$

$2 \mu t = \frac{\lambda}{2}$

$t = \frac{\lambda}{4\mu}$

here we have

$\lambda = 675 nm$

t = 121 nm

now from above equation we have

$\mu = \frac{675 nm}{4(121 nm)}$

$\mu = 1.39$