Question

The diagram below shows the path of a light ray in three different materials. The index of refraction for each material is shown in the upper right portion of the material is shown in the upper right portion of the material Rank the index of refraction values, greatest first.

• A. $n_1,n_2,n_3$
• B. $n_1,n_3,n_2$
• C. $n_2,n_1,n_3$
• D. $n_2,n_3,n_1$
• E. $n_3,n_1,n_2$

Answer 1

The correct option is B $n_1,n_3,n_2$

By Snell's law,

when light ray enters from medium 1 to 2 ,

$\sin i_1/\sin r_1=n_2/n_1$ .....................eq1

when light ray enters from medium 2 to 3 ,

$\sin i_2/\sin r_2=\sin r_1/\sin r_2=n_3/n_2$, ......................eq2

angle of refraction $r_1$ and angle of incidence$i_2$ in medium 2 are equal as they are corresponding angles ,

by multiplying eq1 and eq2 we get ,

$\sin i_1/\sin r_2=n_3/n_1$

now we can see from diagram that $i_1<r_2$

therefore $\sin i_1<sin{r}_2$ and $n_3<n_1$ ,

We know that when a light ray is refracted from a rarer medium to a denser medium , it bends towards the normal and vice-versa .

As the light ray enters from medium 1 to 2 it is going away from normal therefore medium 1 is denser than medium 2 i.e. $n_1>n_2$ Now ray enters in medium 3 from 2 and is bending toward the normal so medium 3 is denser than medium 2 i.e. $n_2<n_3$ ,

as a conclusion we can say $n_1>n_3>n_2$