Final answer:
To determine the index of refraction for glass when light emerges parallel to the surface of water, one would use Snell's Law and the knowledge that the refracted ray at an angle of 90° indicates it is at the critical angle. The index for water (μw = 1.333) is used in the equation μg = (μw)·sin(ic) to find μg, the index of refraction for glass.
Step-by-step explanation:
The question pertains to the phenomenon of light refraction at the interface between glass and water. To find the index of refraction (μg) for glass, when a ray of light emerges parallel to the surface of water, we use Snell's Law which is n1·sin(i) = n2·sin(r). Here, n1 is the index of refraction for glass, sin(i) is the sine of the angle of incidence in glass, n2 is the index of refraction for water and sin(r) is the sine of the angle of refraction in water.
If the light emerges parallel to the surface of the water, the angle of refraction in water would be 90° (since it is parallel), and sin(r) will be 1. The angle of incidence 'i' in the glass would correspond to the critical angle for glass-water interface. If we know the index of refraction for water (which is typically 1.333), we can find out μg for glass using the rearranged form of Snell's Law: μg = (μw)·sin(ic), where μw is the index of water and ic is the critical angle.
To determine the value of sin(ic), we set up the equation using the knowledge that the refracted ray is parallel to the boundary which means r = 90°. Therefore, sin(ic) = μw. Plugging in the known values would give us the index of refraction for glass.