Light refraction problem in water & air

In summary, the problem involves finding the length of the shadow of a 2.00m long pole in a swimming pool when sunlight is incident at 55.0° above the horizon. Using the formula sin(θ2) = (n1/n2)*sin(θ1), where n1=1 for air and n2=1.33 for water, the angle of refraction is found to be 25.55°. However, a mistake is made in calculating the length of the shadow, as it should be the adjacent side (0.35m) rather than the opposite side (0.71m) of the 55° angle. The correct answer is found to be 1.07m.
  • #1
MrMoose
23
0

Homework Statement



In the figure attached, a 2.00m long vertical pole extends from the bottom of a swimming pool to a point 50.0cm above the water. Sunlight is incident at 55.0° above the horizon. What is the length of the shadow of the pole on the level bottom of the pool?

Homework Equations



See figure attached for my drawing.

n1 = 1 for air

n2 = 1.33 for water

sin(θ2) = (n1/n2)*sin(θ1)

The Attempt at a Solution



Since N2 > N1, we know that the beam bends towards the normal.

θ1 = 90° - 55° = 35°

θ2 = arcsin[(1/1.33)*sin(35°)) = 25.55°

From here, it's geometry:

Tan(55°) = X1 / 50cm

X1 = 0.71m

Tan(25.55°) = X2/1.5m

X2 = 0.71m

X = X1 + X2 = 1.43m

According to the back of the book, this is not correct. Where am I going wrong? This seems like a really straight forward problem. Thanks in advance, MrMoose
 

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  • #2
MrMoose said:
From here, it's geometry:

Tan(55°) = X1 / 50cm
This is not the correct angle.
 
  • #3
Hi mfb, please elaborate, I still don't see the error. Tangent of the angle is equal to the length of the opposite side over the length of the adjacent side.
 
  • #4
The opposite side of your 55°-angle is the .5m of the pole, and not the length of the shadow. If you want to calculate this part in the same way as you did the part under water, you need the corresponding angle (relative to the vertical, 35°).
 
  • #5
Oh wow, that was a silly mistake. So...

Tan(55°) = 50cm/X1

X1 = 0.35m

And X = X1 + X2 = 1.07m, which is the correct answer.

Thanks so much for your help.
 

Related to Light refraction problem in water & air

What is light refraction?

Light refraction is the bending of light as it passes through different mediums, such as water and air. This is due to the difference in density between the two mediums, causing the light to change direction.

Why does light refract in water and air?

Light refracts because the speed of light changes as it passes through different mediums. In water, light travels slower than in air, causing it to bend towards the normal line. In air, light travels faster, causing it to bend away from the normal line.

What is the law of refraction?

The law of refraction, also known as Snell's law, states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speeds of light in the two mediums. This can be expressed as n1sinθ1 = n2sinθ2, where n1 and n2 are the refractive indices of the two mediums, and θ1 and θ2 are the angles of incidence and refraction, respectively.

How does the angle of incidence affect light refraction?

The angle of incidence plays a crucial role in light refraction. As the angle of incidence increases, the angle of refraction also increases. However, if the angle of incidence is too large, the light may not pass through the medium and instead be reflected back.

What is the difference between light refraction in water and air?

The main difference between light refraction in water and air is the refractive index of the two mediums. Water has a higher refractive index than air, meaning that light bends more when passing through water compared to air. This is why objects appear closer and larger when viewed through water, such as in a swimming pool or aquarium.

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