To determine the critical angle of a substance for light moving from the substance into air, we need to use Snell's Law and the concept of total internal reflection.
First, let's define some terms:
- [tex]\( n_1 \)[/tex]: The refractive index of the substance (in this case, 1.81).
- [tex]\( n_2 \)[/tex]: The refractive index of air (approximately 1.0).
- [tex]\( \theta_c \)[/tex]: The critical angle, which is the angle of incidence in the substance at which the angle of refraction in the air is 90 degrees.
Snell's Law states:
[tex]\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \][/tex]
At the critical angle, [tex]\( \theta_2 \)[/tex] (the angle of refraction) is 90 degrees, and [tex]\( \sin(90^\circ) = 1 \)[/tex]. So, Snell's Law at the critical angle becomes:
[tex]\[ n_1 \sin(\theta_c) = n_2 \sin(90^\circ) \][/tex]
[tex]\[ n_1 \sin(\theta_c) = n_2 \][/tex]
[tex]\[ \sin(\theta_c) = \frac{n_2}{n_1} \][/tex]
Substitute the given values [tex]\( n_1 = 1.81 \)[/tex] and [tex]\( n_2 = 1 \)[/tex]:
[tex]\[ \sin(\theta_c) = \frac{1}{1.81} \][/tex]
Now, calculate [tex]\( \sin(\theta_c) \)[/tex]:
[tex]\[ \sin(\theta_c) \approx 0.5525 \][/tex]
To find [tex]\( \theta_c \)[/tex], take the inverse sine (arcsin) of 0.5525:
[tex]\[ \theta_c = \arcsin(0.5525) \][/tex]
Using a calculator, we find:
[tex]\[ \theta_c \approx 33.5^\circ \][/tex]
Therefore, the critical angle of the substance with a refractive index of 1.81 as it moves into air is approximately:
[tex]\[ \boxed{33.5^\circ} \][/tex]
