Chapter 33: Q55P (page 1005)

In Fig. 33-55, a $2 .00 m$ long vertical pole extends from the bottom of a swimming pool to a point $50 .0 cm$ above the water. Sunlight is incident at angle $θ = 55 .0 °$ . What is the length of the shadow of the pole on the level bottom of the pool?

Short Answer

Expert verified

The length of the shadow of the pole on the level bottom of the pool is $1.07 m$

Step by step solution

Given

Angle of incidence $θ = 55^{0}$

$l_{1} = 0.50 m l_{2} = 1.50 m$

Understanding the concept

We can use Snell’s law to find the angle of refraction. Then using the geometry of the figure, we can find the length of the shadow of the pole on the level bottom of the pool.

Formula:

$n_{1} \sin θ_{1} = n_{2} \sin θ_{2}$

Calculate the length of the shadow of the pole on the level bottom of the pool

Ray diagram for the given problem:

According to Snell’s law,

$n_{1} \sin θ_{1} = n_{2} \sin θ_{2}$

But, here

$θ_{1} = 90 - θ = 90^{0} - 55^{0} = 35^{0}$

$\begin{matrix} θ_{2} = \sin^{- 1} (\frac{n_{1} \sin θ_{1}}{n_{2}}) \\ θ_{2} = \sin^{- 1} (\frac{(1.0) \sin 35^{0}}{1.33}) \\ θ_{2} = {25.55}^{0} \end{matrix}$

Now, from the figure,

$\begin{matrix} x = l_{1} \tan θ_{1} \\ x = (0.50 m) \tan 35^{0} \\ x = 0.35 m \end{matrix}$

And,

$\begin{matrix} L = l_{2} \tan θ_{2} \\ L = (1.50 m) \tan {25.55}^{0} \\ L = 0.717 \\ ~ 0.72 m \end{matrix}$

Therefore, the total length of the shadow is

$\begin{matrix} x + L = 0.35 m + 0.72 m \\ x + L = 1.07 m \end{matrix}$

Therefore, the length of the shadow of the pole on the level bottom of the pool is $1.07 m$

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In Fig. 33-55, a $2 .00 m$ long vertical pole extends from the bottom of a swimming pool to a point $50 .0 cm$ above the water. Sunlight is incident at angle $θ = 55 .0 °$ . What is the length of the shadow of the pole on the level bottom of the pool?

Short Answer

Step by step solution

Given

Understanding the concept

Calculate the length of the shadow of the pole on the level bottom of the pool

One App. One Place for Learning.

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In Fig. 33-55, a 2.00 mlong vertical pole extends from the bottom of a swimming pool to a point 50.0cmabove the water. Sunlight is incident at angle θ=55.0°. What is the length of the shadow of the pole on the level bottom of the pool?

Short Answer

Step by step solution

Given

Understanding the concept

Calculate the length of the shadow of the pole on the level bottom of the pool

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Engineering Physics

Translational Dynamics

Physical Quantities And Units

Medical Physics

Space Physics

Kinematics Physics

Study anywhere. Anytime. Across all devices.

In Fig. 33-55, a $2 .00 m$ long vertical pole extends from the bottom of a swimming pool to a point $50 .0 cm$ above the water. Sunlight is incident at angle $θ = 55 .0 °$ . What is the length of the shadow of the pole on the level bottom of the pool?