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1. Consider the radio waves from 88.1 MHz KDHX radio.

If you were flying toward the radio station’s broadcast center at 500 mph, by how much (and in which direction) would the broadcast frequency that you receive change?
2. Given that the sun delivers about 1.36 kW per square meter to earth orbit, how much power from a direct look at the sun might be incident on your eye’s 2.2mm diameter "bright-adapted" pupil? How might this suggest why laser pointers are limited to 5mW?

1 Answer

6 votes

Answer and Explanation:

1. If you were flying toward the broadcast center of KDHX radio at 500 mph, the broadcast frequency that you receive would increase. This phenomenon is known as the Doppler effect. As you approach the source of the radio waves, the frequency appears higher.

To calculate the change in frequency, we can use the formula:

Δf/f = (v/c) where Δf is the change in frequency, f is the original frequency, v is the velocity of the observer (in this case, 500 mph), and c is the speed of light.

Given that the original frequency is 88.1 MHz and the speed of light is approximately 3 x 10^8 meters per second, we can calculate the change in frequency as follows:

Δf/88.1 MHz = (500 mph / (3 x 10^8 m/s))

Δf = 88.1 MHz * (500 mph / (3 x 10^8 m/s))

Δf ≈ 0.147 kHz

Therefore, the broadcast frequency that you receive would increase by approximately 0.147 kHz as you fly toward the broadcast center.

2. The power delivered by the sun to Earth orbit is approximately 1.36 kW per square meter. If we consider the direct look at the sun through the eye's 2.2mm diameter "bright-adapted" pupil, we can calculate the power incident on the pupil.

The area of the pupil can be calculated using the formula for the area of a circle:

A = πr^2

A = π(1.1 mm)^2

A ≈ 3.801 mm^2

Converting this area to square meters:

A ≈ 3.801 x 10^-6 m^2

To calculate the power incident on the pupil, we multiply the solar power density by the area:

Power = (1.36 kW/m^2) * (3.801 x 10^-6 m^2)

Power ≈ 5.172 x 10^-3 W

Therefore, the power incident on the eye's pupil would be approximately 5.172 milliwatts (mW).

This suggests why laser pointers are limited to 5mW. Direct exposure to laser light, even at low powers, can be harmful to the eyes. By limiting laser pointers to 5mW, there is a reduced risk of potential eye damage. It is important to follow safety guidelines and avoid looking directly into laser beams to protect the eyes from potential harm.