
Light travels in straight lines at 3×10^{8} m/s. Light carries energy and information and can vary in intensity. The intensity of light has units of power divided by area, or watts per square meter. Light has both wave and particle properties, and its energy determines its color. Light rays can be diverted when they interact with matter. The properties of light explain the formation of shadows, why Earth has cycles of night and day, why there are eclipses of the Sun and Moon, and why there are phases of the Moon. Light intensity decreases with distance following an inverse square law. Gravitational and electrical forces are other examples of inverse square laws.

ray diagram, inverse square law, object distance, image distance



Review problems and questions 

 The distance between the Earth and the Sun is one astronomical unit (AU), which is 1.50×10^{8} km. A light year (ly) is the distance light travels in one year.
 How many minutes does it take light to travel from the Sun to the Earth?
 How many kilometers is a light year?
 How many AU is a light year?

Answer:  8 min, 20 s
 9.46×10^{12} km
 63,072 AU
Solution: Asked: time t for light to travel from the Sun to the Earth
Given: distance traveled, d = 1 AU =1.50×10^{8} km Relationships: v = d/t; 1 km = 1000 m; 60 s = 1 min; speed of light, c = 3×10^{8} m/s Solve:$$v=\frac{d}{t}$$Rearrange equation and solve for t:$$t=\frac{d}{v}=\frac{\left(1.50\times {10}^{8}\text{km}\right)\left(\frac{\mathrm{1,000}\text{m}}{1\text{km}}\right)}{3\times {10}^{8}\text{m/s}}=500\text{s}$$Convert to minutes:$$\left(500\text{s}\right)\left(\frac{1\text{min}}{60\text{s}}\right)=8.33\text{min}$$Answer: 8 min and 20 s  Asked: kilometers in one light year
Given: light traveling at the speed of light for one year, t = 1 yr Relationships: v = d/t; speed of light c = 3×10^{8} m/s; 1,000 m = 1 km Solve: Calculate number of seconds in a year:$$\left(1\text{yr}\right)\left(\frac{365\text{days}}{1\text{yr}}\right)\left(\frac{24\text{hr}}{1\text{day}}\right)\left(\frac{60\text{min}}{1\text{hr}}\right)\left(\frac{60\text{s}}{1\text{min}}\right)=\mathrm{31,536,000}\text{s}$$Rearrange and solve v = d/t for d:$$d=vt=\left(3\times {10}^{8}\text{m/s}\right)\left(\mathrm{31,536,000}\text{s}\right)=9.4608\times {10}^{15}\text{m}$$Convert to kilometers:$$\left(9.4608\times {10}^{15}\text{m}\right)\left(\frac{1\text{km}}{\mathrm{1,000}\text{m}}\right)=9.4608\times {10}^{12}\text{km}$$Answer: 9.4608×10^{12} km
 Convert previous answer to astronomical units:$$\left(9.4608\times {10}^{12}\text{km}\right)\left(\frac{1\text{AU}}{1.50\times {10}^{8}\text{km}}\right)=\mathrm{63,072}\text{AU}$$Answer: 63,072 AU

 What is the difference between a solar and a lunar eclipse?
 Which body’s shadow projects onto another in a solar eclipse? In a lunar eclipse?
 Which shadow is larger? Why?

 In a solar eclipse, the Moon is between the Sun and the Earth, so the view of the Sun is obstructed on some parts of Earth’s surface. In a lunar eclipse, the Earth is between the Sun and the Moon, so no light is cast on the Moon and it cannot be seen on some parts of Earth's surface.
 In a solar eclipse, the shadow of the Moon is projected onto the Earth. In a lunar eclipse, the shadow of the Earth is projected onto the Moon.
 Because the Earth is much larger than the Moon, the shadow cast on the Moon by the Earth during a lunar eclipse is much larger. This means that a lunar eclipse is visible from more places on Earth.

 The dwarf planet Makemake has an average orbital distance of 45.8 AU from the Sun. If sunlight reaching the Earth has an intensity of 1,365 W/m^{2}, what is the intensity of sunlight reaching Makemake?

Answer: 0.651 W/m^{2} The intensity of light falls off as 1/r^{2}, so use the square of the ratio of the orbital distances between Earth (1 AU) and Makemake (45.8 AU) to determine the intensity of the light reaching the latter:$${I}_{Makemake}={I}_{Earth}{\left(\frac{{r}_{Earth}}{{r}_{Makemake}}\right)}^{2}=\left(1,365{\text{W/m}}^{2}\right){\left(\frac{1\text{AU}}{45.8\text{AU}}\right)}^{2}=0.651{\text{W/m}}^{2}$$

 Where is the lens of a pinhole camera located?

There is no lens! A pinhole camera creates images using the property that light travels in straight lines. It does not bend the path of light, which is what lenses do!

 The light intensity at a spot 1.0 m away from a bright lightbulb is 30 W/m^{2}.
 What is the intensity at a distance of 2.0 m?
 What is the intensity at a distance of 0.5 m?
 At what distance from the bulb will the intensity be 15 W/m^{2}?

Answer:  7.5 W/m^{2}
 120 W/m^{2}
 1.4 m

 Sometimes near midday on a sunny day, you can see many images of the Sun projected onto the ground underneath a leafy tree. What creates the images?

If there are tiny gaps between the leaves, each little gap acts as a pinhole camera and projects an image of the Sun onto the ground. Many little gaps between the leaves produce many pinhole camera images of the Sun!

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