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A ray of light traveling in water is incident at a water-air interface. The angle of incidence is slowly increased until total internal reflection first occurs. If the refractive index of water is 1.333, then the angle of incidence must have been ______.

Respuesta :

ajeigbeibraheem

Answer:

45.72°

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  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 ⁻¹(0.7159)

i = 45.72°

Thus, If the refractive index of water is 1.333, then the angle of incidence must have been 45.72°

onyebuchinnaji

Answer:

The incident angle is 48.6⁰

Explanation:

Given;

the refractive index of water, η = 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]

η = 1/sinθc

sinθc = 1/η

sinθc = 1/1.333

sinθc = 0.7502

θc = sin⁻¹(0.7502)

θc = 48.6⁰

Thus, the incident angle is 48.6⁰