n=1.331 for red light in water and n=1.34 for blue light in water. If a ray of white light enters the water at 83 degrees what are the underwater angles of refraction for the blue and red components of white light?

Also using Snell's law :
(sin theda 1/ sin theda 2) = (n2/n1)
so that theda 1 =83 degrees. What would be n2 and n1? Is n2 for the blue or red n also what would n1 be?

To find the underwater angles of refraction for the blue and red components of white light, we can use Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media.

In this case, as stated, the refractive index (n) values for red light in water are n=1.331, and for blue light in water, n=1.34.

Using Snell's law: (sin(theta1) / sin(theta2)) = (n2 / n1)

Given that theta1 is 83 degrees, we want to find n2 and n1 for the blue and red components of white light.

Since red light has n=1.331 for water, we can assign n1=1.331 for red light.

Similarly, since blue light has n=1.34 for water, we can assign n2=1.34 for blue light.

Now we can use Snell's law to find the underwater angles of refraction for the blue and red components of white light.

For the blue component:
(sin(theta1) / sin(theta2_blue)) = (n2_blue / n1_blue)

Substituting the known values:
(sin(83) / sin(theta2_blue)) = (1.34 / 1.331)

Next, we can solve for theta2_blue by rearranging the equation:
sin(theta2_blue) = sin(83) * (1.331 / 1.34)
theta2_blue = sin^(-1)(sin(83) * (1.331 / 1.34))

For the red component:
(sin(theta1) / sin(theta2_red)) = (n2_red / n1_red)

Substituting the known values:
(sin(83) / sin(theta2_red)) = (1.34 / 1.331)

Next, we can solve for theta2_red by rearranging the equation:
sin(theta2_red) = sin(83) * (1.331 / 1.34)
theta2_red = sin^(-1)(sin(83) * (1.331 / 1.34))

By evaluating the above expressions, you will calculate the underwater angles of refraction for the blue and red components of white light.