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A compact disc (CD) is read from the bottom by a semiconductor laser beam with a wavelength of 790 nm that passes through a plastic substrate of refractive index 1.80. When the beam encounters a pit, part of the beam is reflected from the pit and part from the flat region between the pits, so these two beams interfere with each other.What must the minimum pit depth be so that the part of the beam reflected from a pit cancels the part of the beam reflected from the flat region? (It is this cancellation that allows the player to recognize the beginning and end of a pit.)

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okpalawalter8

Answer:

The value is   [tex]t  = 110 nm[/tex]

Explanation:

From the question we are told that  

   The wavelength of the beam is [tex]\lambda  =  790 \  nm  =  790 *10^{-9} \  m[/tex]

   The refractive index is  [tex]n_r  =  1.80[/tex]

Generally from the condition for destructive interference  the depth  of the pit  is mathematically represented as

     [tex]t  = [ m  + \frac{1}{2} ] *  \frac{\lambda}{n_r}  *  \frac{1}{2}[/tex]

Here m which is the order of the fringe is zero because both beams cancel out

So

       [tex]t  = [ 0  + \frac{1}{2} ] *  \frac{790 *10^{-9}}{ 1.80}  *  \frac{1}{2}[/tex]

=>     [tex]t  = 110 *10^{-9} \  m[/tex]

=>     [tex]t  = 110 nm[/tex]

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