Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle α can have without any light being refracted out of the prism at face AC if the prism is immersed in water (with index of refraction 1.33). Express your answer in degrees. Ignore any reflections from the surface BC.

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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