Answer:
45.72Ā°
Explanation:
A ray of light traveling in water is incident at a water-air interface.
Let the angle of incidence be = i which is relative to Ā a right angle
refractive index of water [tex]\mu_l[/tex] = 1.333
refractive index [tex]\mu_2= 1[/tex]
By applying snell's law of refraction
[tex]\mu_l sin\ i = \mu _2 sin 90^0[/tex]
1.33 sin i = 1 (1)
sin i = [tex]\frac{1}{1.33}[/tex]
sin i = 0.7159
i = sin ā»Ā¹(0.7159)
i = 45.72Ā°
Thus, If the refractive index of water is 1.333, then the angle of incidence must have been 45.72Ā°
Answer:
The incident angle is 48.6ā°
Explanation:
Given;
the refractive index of water, Ī· = 1.333
Total internal reflection occurs when incident light travels from more optically dense medium towards less optically dense medium.
Example water to air;
During this process incident angle equals critical angle of incident;
[tex]\theta_i = \theta_c[/tex]
Ī· = 1/sinĪøc
sinĪøc = 1/Ī·
sinĪøc = 1/1.333
sinĪøc = 0.7502
Īøc = sinā»Ā¹(0.7502)
Īøc = 48.6ā°
Thus, the incident angle is 48.6ā°
