REFRACTION THROUGH A COMPOUND SLAB

Home » Ray optics and optical instruments » REFRACTION THROUGH A COMPOUND SLAB

REFRACTION THROUGH A COMPOUND SLAB

Posted on

clip_image002

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}}