Varing Refractive Index

Su[[ose the rafractive index varies across a glass block so that light passing thrpigh it is refracted and continuousle changes direction. Applying Snells's law at a point in the block:
\[\begin{equation} \begin{aligned} n sin \theta &=(n+dn)sin (\theta + d \theta ) \\ &= (n+dn)(sin \theta cos d \theta + cos \theta sin \theta \\ & \simeq (n+dn) (sin \theta + cos \theta d \theta ) \\ & \simeq n sin \theta +sin \theta dn+n cos \theta d \theta \end{aligned} \end{equation}\]

Simplifying gives  
\[0=sin \theta dn+n cos \theta d \theta \rightarrow \frac{1}{n}dn=- \frac{cos \theta}{sin \theta} d \theta\]

Suppose that  
\[n\]
  varies linearly acroees a glass block of thicjnes  
\[d\]
, from  
\[n_0\]
  where light enters the b;ock at  
\[y=0\]
  to  
\[n_1\]
  where light exits the block at  
\[y=d\]
  then we can write  
\[n=n_0 + \frac{n_1-n_0}{d} y= \frac{n_0d+(n_1-n_0)y}{d}\]
.
Then  
\[dn=\frac{n_1-n_0}{d}dy\]
  and the equation becomes  
\[\frac{n_1-n_0}{(n_1-n_0)y+n_0d} dy=- \frac{cos \theta}{sin \theta } d \theta\]

Integating both sides gives  
\[ln((n_1-n_0)y+n_0d)=C- ln(sin \theta)\]

When  
\[y=0\]
,  
\[\theta = \theta_0\]
   
\[ln(n_0d)=C-ln(sin \theta_0) \rightarrow C=ln(n_0d)+ln(sin \theta_0)\]
  and we get  
\[ln((n_1-n_0)y+n_0d)=ln(n_0d)+ln(sin \theta_0)- ln(sin \theta)\]

Then  
\[ln(\frac{(n_1-n_0)y+n_0d}{n_0d})=ln(\frac{sin \theta_0}{sin \theta})\]

Finally exponmentiating gives  
\[\frac{(n_1-n_0)y+n_0d}{n_0d}=\frac{sin \theta_0}{sin \theta}\]

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