A fish swimming in a pond looks up at a dragonfly flying 45 cm above the surface of the water. How far does the dragonfly appear to the fish if the indices of refraction of water and air are 1.33 and 1, respectively? Enter your answer in cm.
To determine how far the dragonfly appears to the fish, we need to consider the refraction of light as it passes from air to water.
The difference in refraction between two mediums can be determined by Snell's Law, which states:
n1 * sin(theta1) = n2 * sin(theta2)
Here:
- n1 is the index of refraction of the first medium (air)
- n2 is the index of refraction of the second medium (water)
- theta1 is the angle of incidence of the light ray in the first medium
- theta2 is the angle of refraction of the light ray in the second medium
Let's assume the light ray travels from the dragonfly to the fish's eyes. We can consider the incident ray as coming vertically down from the dragonfly and hitting the water surface. The angle of incidence (theta1) will be 90 degrees.
Using Snell's Law, we can rearrange the equation to solve for sin(theta2):
sin(theta2) = (n1 / n2) * sin(theta1) = (1 / 1.33) * sin(90)
sin(theta2) = 0.7518796992
Now, to find the apparent distance of the dragonfly to the fish, we can use trigonometry. We can consider a right-angled triangle formed by the fish's line of sight, the water surface, and the apparent position of the dragonfly. The angle theta2 corresponds to the angle between the fish's line of sight and the water surface.
Using the sine function:
sin(theta2) = height of the dragonfly / apparent distance
Since the height of the dragonfly is given as 45 cm, we can solve for the apparent distance:
apparent distance = height of the dragonfly / sin(theta2)
apparent distance = 45 / 0.7518796992
Therefore, the dragonfly appears to be approximately 59.85 cm away from the fish.