Question regarding light and reflection!

Question

anonymous · Answer

attachment

anonymous · Answer

@ganeshie8 Hey Ganesh! Do you know much about this material in particular?

ganeshie8 · Answer

sorry no idea, let me tag @Vincent-Lyon.Fr

anonymous · Answer

No problem X)

anonymous · Answer

I think I figured it out but I might be wrong...

We have the initial equation for destructive interference:$$\huge \phi=\frac{4\pi n_b t \cos(\theta_i)}{\lambda}-\phi_{r2}-\phi_{r1}$$Since we know that the speed will be greater in air than in glass ( $c_2<c_1)$, then there is a phase shift of $\pi$. Because $\phi_{r1} = \phi_{r2}$, then the equation simplifies to:$$\huge 1=\frac{4n_b t \cos(\theta_i)}{\lambda}$$Additionally, we are given that the incidence is normal, which means that $\theta_i=0 \implies \cos(0)=1$. Now we have$$\huge 1=\frac{4n_b t}{\lambda}$$Plugging in what we know and solving for $n_b$, we then have$$\huge n_b=\frac{550\text{nm}}{4(104\text{nm})}=1.32\text{nm}$$

I am not 100% certain, though..

anonymous · Answer

@Astrophysics

anonymous · Answer

Oops, the initial equation is supposed to have

$\huge +~\phi_{r2}$

vincent-lyon.fr · Answer

Hi, no time now to give a full answer, but you can play with this (French) applet that will help you understand what is at stake. http://ressources.univ-lemans.fr/AccesLibre/UM/Pedago/physique/02/optiphy/antirefl.html