REFRACTION THROUGH A COMPOUND SLAB

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Consider two parallel slabs having refractive index {n_b}and{n_c}are kept one above the other. The medium above and below the compound slab is the same.

If we consider refraction of light at the interface of ‘a’ and ‘b’, then

\frac{{\sin i}}{{\sin r}} = {n_{b\,a}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..................\,\left( 1 \right)

Now consider refraction of light at the interface of ‘b’ and ‘c’, then

\frac{{\sin r}}{{\sin r'}}\, = {n_{c\,b}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..................\,\left( 2 \right)

Considering refraction of light at the interface of c and a, we get

\frac{{\sin r'}}{{\sin i}} = {n_{a\,c}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..................\,\left( 3 \right)

Multiplying equations (1), (2) and (3), we get

{n_{b\,a}} \times {n_{c\,b}} \times {n_{a\,c}} = 1or {n_{b\,a}} \times {n_{c\,b}} = \frac{1}{{{n_{a\,c}}}} or {n_{b\,a}} \times {n_{c\,b}} = {n_{c\,a}}

{n_{c\,a}} = {n_{c\,b\,}} \times \,{n_{b\,a}}

 

Post Author: E-Physics

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