What is the angle of refraction (relative to the boundary normal) of a diamond in contact with water according to Snell's law? The indices of refraction for water and diamond are respectively $1.33$ and $2.42$ .

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Answer 1 · 2015-07-24T23:28:32

The equation for this situation is

$n_"water"sintheta_"water" = n_"diamond"sintheta_"diamond"$ .

Since we know $theta_"water"$ , and $n_"water" ~~ 1.33$ and $n_"diamond" ~~ 2.42$ , solve for $theta_"diamond"$ .

$theta_"diamond" = arcsin(n_"water"/n_"diamond" sintheta_"water")$

Remember that angles of refraction are with respect to the normal.

What is the angle of refraction (relative to the boundary normal) of a diamond in contact with water according to Snell's law? The indices of refraction for water and diamond are respectively 1.331.33 and 2.422.42.