Question

Constants Part A You (height of your eyes above the water, h 1.55 m) are standing 2.00 m from the edge of a 2.50-m-deep swimming pool (Figure 1). You notice that you can barely see your cell phone, which went missing a few minutes before, on the bottom of the pool. The index of refraction of water at this temperature is 1.333 How far from the side of the pool is your cell phone? Express your answer with the appropriate units to three significant figures. d Value Units Submit Request Answer Provide Feedback Figure 1 of 1 2.00 m 2.50 m

Answer 1

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From the figure we can see that Angle $\theta$ will be:

tan $\theta$ = h/2.00 = 1.55/2.00

$\theta$ = arctan (1.55/2.00) = 37.77 deg

So Angle of incident A1 will be

A1 = 90 - $\theta$ = 90 - 37.77 = 52.23 deg

n1 = refractive index of air = 1.00

n2 = refractive index of water = 1.333

A2 = Angle of refraction = ?

Using Snell's law:

n1*sin A1 = n2*sin A2

A2 = arcsin (n1*sin A1/n2)

A2 = arcsin ((1*sin 52.23deg)/1.333)

A2 = 36.37 deg

Now in lower right-angle triangle

tan A2 = d/2.50

d = 2.50*tan 36.37 deg

d = 1.84 m

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