Answer to Question #103118 in Optics for DD

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

As per the given data in the question,

Refractive index of the film=1.34

thickness of the extreme side 0 and t

wavelength of the light (λ)=492nm(\lambda)=492nm

(λ)=492nm

Number of fringes=20

As per the rule of x-ray reflection on the single layer

$2t\sqrt{n^2−\sin^2i}-\dfrac{\lambda}{2}​=mλ$

n=refractive index of the medium

Light is getting incident normally, so i=0i=0

i=0

So,

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

$\Rightarrow t=\dfrac{m\lambda+\dfrac{\lambda}{2}}{2n}$

$\Rightarrow t=\dfrac{20\times492\times10^{-9}+246\times 10^{-9}}{2\times 1.34}$

$\Rightarrow t=9931.8\times10^{-9}m$

$\Rightarrow t=9.9318\times 10^{-6}m$

$t=9.93\mu m$