Question

A non-reflective coating that has a thickness of 198 nm (n = 1.45) is deposited on top of a substrate of glass (n = 1.50). What wavelength of visible light is most strongly transmitted if it is illuminated perpendicular to its surface?

Answer 1

Answer:

The  wavelength is $\lambda_ 1 = 574.2 nm$

Explanation:

From the question we are told that  

      The  thickness is $t = 198 nm = 198 *10^{-9 }\ m$

      The refractive  index of the non-reflective coating is  $n_m = 1.45$  

       The  refractive  index of glass is $n_g = 1.50$

       

Generally the condition for  destructive  interference is mathematically represented as

            $2 * n_m * t * cos (\theta) = n * \lambda$

Where $\thata$ $\theta$ is the angle of refraction which is  0° when the light is strongly transmitted

    and  n is the order maximum interference

        so  

             $\lambda = \frac{2 * n * t * cos (\theta )}{n}$

at the point n =  1  

           $\lambda _1 = \frac{2 * 1.45 * 198*10^{-9} * cos (0 )}{1}$

           $\lambda_1 = 574.2 *10^{-9}$

          $\lambda_1 = 574.2 nm$

at  n =2  

         $\lambda _2 = \frac{\lambda _1 }{2}$

         $\lambda _2 = \frac{574.2*10^{-9} }{2}$

         $\lambda _2 = 2.87 1 *10^{-9} \ m$

         $\lambda _2 = 287. 1 nm$

Now we know that the wavelength range of visible light is  between

           $390 \ nm \to 700 \ nm$

   So the wavelength of visible light that is been transmitted is  

          $\lambda_ 1 = 574.2 nm$