A laser pointer is shined through air (n=1.0) and enters another medium at an angle of 45degrees. If the angle of refraction is 24°. What is the index of refraction of the medium.

First, use Snell's Law:

n_1 \sin( \theta_1 ) = n_2 \sin( \theta_2 )

Given:
n_1 = 1.0
\theta_1 = 45^\circ
\theta_2 = 24^\circ

Rearrange Snell's Law to solve for \( n_2 \):

n_2 = \frac{ n_1 \sin( \theta_1 ) }{ \sin( \theta_2 ) }

Substitute the given values:

n_2 = \frac{ 1.0 \sin( 45^\circ ) }{ \sin( 24^\circ ) }

Calculate the sine values:
\sin( 45^\circ ) = 0.7071
\sin( 24^\circ ) = 0.4067

Now plug in the sine values to find \( n_2 \):

n_2 = \frac{ 0.7071 }{ 0.4067 }

n_2 = 1.74

So, the index of refraction of the medium is:

n_2 = 1.74