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A small red light-emitting diode (LED) is placed 15.0 cm below the surface of the water in a bathtub. Because of total internal reflection, a circle of red light is observed at the water surface. Light inside the circle gets out into the air, but light hitting the water surface (from below) outside the circle experiences total internal reflection. Calculate the diameter of this circle. Assume that the air above the water has an index of refraction of 1.00, and that the water has an index of refraction of 1.33. diameter

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Chrisnando

Answer:

34.2 Cm

Explanation:

Lets take, Snell's law

[tex] n_2 sinC = n_1 sin90[/tex]

[tex] SinC = \frac{1*1}{1.33} [/tex]

[tex] C = Sin^-^1 \frac{1}{1.33}[/tex]

C = 48.75°

To find radius,r of the circle, we have:

r = H tanC

Where H = 15 cm

r = 15 * tan048.75°

= 17.1 cm

Diameter = 2r

= 2 * 17.1cm

= 34.2 cm

The diameter of the circle is 34.2cm

Answer:

Diameter of the circle = 34.2 cm

Explanation:

The refractive index of water, i = 1.33

Critical angle, c can be calculated as:

[tex]c = sin^{-1} (\frac{1}{i} )\\c = sin^{-1} (\frac{1}{1.33} )\\c = 48.75^{0}[/tex]

From the diagram attached:

[tex]tan c = \frac{r}{h} \\tan 48.75 = \frac{r}{15}\\r = 15 tan 48.75\\r = 17.1 cm[/tex]

Radius, r = 17.1 cm

Diameter of the circle, d = 2r

d = 2 * 17.1

d = 34.2 cm

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