Respuesta :
To solve this problem it is necessary to use the concepts related to Snell's law.
Snell's law establishes that reflection is subject to
[tex]n_1sin\theta_1 = n_2sin\theta_2[/tex]
Where,
[tex]\theta =[/tex] Angle between the normal surface at the point of contact
n = Indices of refraction for corresponding media
The total internal reflection would then be given by
[tex]n_1 sin\theta_1 = n_2sin\theta_2[/tex]
[tex](1.54) sin\theta_1 = (1.33)sin(90)[/tex]
[tex]sin\theta_1 = \frac{1.33}{1.54}[/tex]
[tex]\theta = sin^{-1}(\frac{1.33}{1.54})[/tex]
[tex]\theta = 59.72\°[/tex]
Therefore the [tex]\alpha_{max}[/tex] would be equal to
[tex]\alpha = 90\°-\theta[/tex]
[tex]\alpha = 90-59.72[/tex]
[tex]\alpha = 30.27\°[/tex]
Therefore the largest value of the angle α is 30.27°
The largest value the angle α can have without any light being refracted out of the prism at face AC is 30.27°.
How to calculate the angle?
From the information given, the total internal reflection would be:
1.54sinb = (1.33) sin90°
sin b = (1.33 sin90° / 1.54)
b = 59.72°
Therefore, the value of the angle will be:
= 90° - 59.72°
= 30.27°
In conclusion, the correct option is 30.27°.
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