Essential Physics

Section 1 review

Light travels in straight lines at 3×10⁸ 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. Read the text aloud

Read the text aloud

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

I = \frac{P}{A}

Review problems and questions

The distance between the Earth and the Sun is one astronomical unit (AU), which is 1.50×10⁸ km. A light year (ly) is the distance light travels in one year.
1. How many minutes does it take light to travel from the Sun to the Earth?
2. How many kilometers is a light year?
3. How many AU is a light year?

Answer:

8 min, 20 s
9.46×10¹² km
63,072 AU

Solution:

Asked: time t for light to travel from the Sun to the Earth
Given: distance traveled, d = 1 AU =1.50×10⁸ km
Relationships: v = d/t; 1 km = 1000 m; 60 s = 1 min; speed of light, c = 3×10⁸ m/s
Solve: $v = \frac{d}{t}$ Rearrange equation and solve for t: $t = \frac{d}{v} = \frac{(1.50 \times 10^{8} km) (\frac{1,000 m}{1 km})}{3 \times 10^{8} m/s} = 500 s$ Convert to minutes: $(500 s) (\frac{1 min}{60 s}) = 8.33 min$ 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 = 3×10⁸ m/s; 1,000 m = 1 km
Solve:
Calculate number of seconds in a year: $(1 yr) (\frac{365 days}{1 yr}) (\frac{24 hr}{1 day}) (\frac{60 min}{1 hr}) (\frac{60 s}{1 min}) = 31,536,000 s$ Rearrange and solve v = d/t for d: $d = v t = (3 \times 10^{8} m/s) (31,536,000 s) = 9.4608 \times 10^{15} m$ Convert to kilometers: $(9.4608 \times 10^{15} m) (\frac{1 km}{1,000 m}) = 9.4608 \times 10^{12} km$ Answer: 9.4608×10¹² km
Convert previous answer to astronomical units: $(9.4608 \times 10^{12} km) (\frac{1 AU}{1.50 \times 10^{8} km}) = 63,072 AU$ Answer: 63,072 AU

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

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/m², what is the intensity of sunlight reaching Makemake?

Answer: 0.651 W/m²
The intensity of light falls off as 1/r², 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_{M a k e m a k e} = I_{E a r t h} {(\frac{r_{E a r t h}}{r_{M a k e m a k e}})}^{2} = (1, 365 {W/m}^{2}) {(\frac{1 AU}{45.8 AU})}^{2} = 0.651 {W/m}^{2}

Where is the lens of a pinhole camera located?

The light intensity at a spot 1.0 m away from a bright lightbulb is 30 W/m².
1. What is the intensity at a distance of 2.0 m?
2. What is the intensity at a distance of 0.5 m?
3. At what distance from the bulb will the intensity be 15 W/m²?

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?


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