Question

The sun is 1.5×1011m from earth; its diameter is 1.4×109m . A hiker who wishes to start a fire uses a 5.0-cm -diameter lens with a focal length of 10 cm to focus the sun's rays onto a piece of wood. On a clear day, the intensity of midday sunlight is about 1000 W/m2 . What is the light intensity at the focus of the lens? Assume that the lens reflects 10 % of the sun's light and transmits 90 % ; these are typical values for a lens that does not have an antireflection coating. Express your answer with the appropriate units.

Answer 1

The light intensity at the focus of the lens is 3600 W/m².

Step-by-step explanation:

To find the light intensity at the focus of the lens, we need to consider the amount of light that passes through the lens. The lens reflects 10% of the sun's light and transmits 90%. The intensity of midday sunlight is 1000 W/m². Since the lens focuses the sun's rays onto a piece of wood, the light is concentrated onto a smaller area. Thus, the light intensity at the focus of the lens will be higher.

We can use the formula:

Intensity = (transmitted intensity) * (concentration factor)

The transmitted intensity is 90% of the incident intensity, so it is 900 W/m². The concentration factor can be calculated by dividing the area of the lens (π * (radius)²) by the area of the focused spot (π * (radius/2)²).

Plugging in the values, we get:

Concentration factor = (π * (0.025 m)²) / (π * (0.0125 m)²) = (0.025 m)² / (0.0125 m)² = 4

Thus, the light intensity at the focus of the lens is:

Intensity = 900 W/m² * 4 = 3600 W/m²

Answer 2

To calculate the light intensity at the focus of the lens, you need to calculate the power of sunlight incident on the lens and the area of the lens.

Step-by-step explanation:

The light intensity at the focus of the lens can be calculated using the formula for intensity, which is power divided by area. In this case, the power is the power of sunlight incident on the lens and the area is the area of the lens.

First, let's calculate the power of sunlight incident on the lens. The intensity of midday sunlight is given as 1000 W/m². The lens has a diameter of 5.0 cm, so its radius is 2.5 cm. The area of the lens is, therefore, π × (2.5 cm)². To convert the area to the appropriate units, we need to convert cm to m by dividing by 100. So the power of sunlight incident on the lens is 1000 W/m² × π × (2.5 cm / 100)².

Next, let's calculate the area of the lens. The diameter of the lens is 5.0 cm, so its radius is 2.5 cm. The area of the lens is, therefore, π × (2.5 cm)².

Finally, we can calculate the light intensity at the focus of the lens by dividing the power by the area. So the light intensity at the focus of the lens is (1000 W/m² × π × (2.5 cm / 100)²) / (π × (2.5 cm)²).