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What is the critical angle between glass [tex]\( \left(n_1=1.50\right) \)[/tex] and water [tex]\( \left(n_2=1.33\right) \)[/tex]?

A. 17 degrees
B. 62 degrees
C. 88 degrees
D. 112 degrees

Answer :

To find the critical angle between glass and water, we'll use the concept of total internal reflection. The critical angle is the angle of incidence for which the angle of refraction is 90 degrees. We can calculate it using Snell's law:

[tex]\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \][/tex]

For the critical angle [tex]\( \theta_c \)[/tex], the angle of refraction [tex]\( \theta_2 \)[/tex] is 90 degrees. The sine of 90 degrees is 1, so the equation becomes:

[tex]\[ n_1 \sin(\theta_c) = n_2 \times 1 \][/tex]

Rearranging to find the critical angle:

[tex]\[ \sin(\theta_c) = \frac{n_2}{n_1} \][/tex]

Given:
- [tex]\( n_1 = 1.50 \)[/tex] (refractive index of glass)
- [tex]\( n_2 = 1.33 \)[/tex] (refractive index of water)

Substitute the values into the equation:

[tex]\[ \sin(\theta_c) = \frac{1.33}{1.50} \][/tex]

Calculate [tex]\( \sin(\theta_c) \)[/tex]:

[tex]\[ \sin(\theta_c) = 0.8867 \][/tex]

To find [tex]\( \theta_c \)[/tex], take the inverse sine (arcsin) of 0.8867:

[tex]\[ \theta_c \approx 62 \text{ degrees} \][/tex]

Thus, the critical angle between glass and water is approximately 62 degrees. The correct answer is 62 degrees.

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