Question

4. The intensity of light is proportionate to the inverse square of the distance to the...

4. The intensity of light is proportionate to the inverse square of the distance to the light source. At the same distance, Lightbulb A shines at 9 times the intensity as Lightbulb B. How many feet do you need to stand from Lightbulb B so it appears to have the same intensity as Lightbulb A at 12 feet away.

Science   Physics   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Intensity of light is inversely proportional to the square of the distance to the light source

K is the constant of proportionality.

For lightbulb A

For lightbulb B

At the same distance lightbulb A shines 9 times the intensity as lightbulb B

IA = 9IB

Therefore KA = 9KB

Let dB be the distance to stand from lightbulb B so it appears to have the same intensity as lightbulb A at 12 feet away

Distance of lightbulb A = dA = 12 feet

Distance of lightbulb B = dB

At this point it appears that both lightbulbs have same intensity

IA = IB

dB2 = 144/9 = 16

dB = 4 feet

You need to stand 4 feet away from lightbulb B so it appears to have the same intensity as lightbulb A at 12 feet away.

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The intensity ,I from a point source decreases with the inverse square of the distance,r from...
The intensity ,I from a point source decreases with the inverse square of the distance,r from the source.Suppose that you are listening to your favourite music coming away from a speaker(regarded as a point source and in an open area) placed exactly r1 from you.then you suddenly decide to move away to a new position ,r2 which is exactly twice as far away from the speaker. Calculate by how many decibels ,dB does the sound intensity level ,β drop ,when...
The light intensity a certain distance away from a spherically-symmetric light source is 1200 W/m2. What...
The light intensity a certain distance away from a spherically-symmetric light source is 1200 W/m2. What would be the new intensity (in W/m2) if the source power were tripled while the observation point is made to be five times farther from the source? A. 156 B. 975 C. 144 D. 245 E. None of these
What is the relationship between intensity and distance from a point source? Why should this be...
What is the relationship between intensity and distance from a point source? Why should this be the cast? (Hint: think of the point source as being at the center of the sphere. How are the area of this sphere and the intensity of light related? It might help you to realize that no matter how far away you are the toal power output from the point source is the same. In other words, if you enclose a 50W light bulb...
The illumination of an object by a light source is directly proportional to the strength of...
The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one four times as strong as the other, are placed 12 ft apart, how far away from the stronger light source should an object be placed on the line between the two sources so as to receive the least illumination? (Round your answer to two decimal...
Alpha Centauri A lies at a distance of 4.4 light-years and has an apparent brightness in...
Alpha Centauri A lies at a distance of 4.4 light-years and has an apparent brightness in our night sky of 2.7×10−8 2.7 × 10 − 8 watt/m2 w a t t / m 2 . Recall that 1light−year=9.5×1012km=9.5×1015m 1 l i g h t − y e a r = 9.5 × 10 12 k m = 9.5 × 10 15 m. Suppose you have a light bulb that emits 200 watts w a t t s of visible light....
4. Electrostatics Coulomb’s law, or Coulomb’s inverse-square law, is a law of physics that describes force...
4. Electrostatics Coulomb’s law, or Coulomb’s inverse-square law, is a law of physics that describes force interactions between static electrically charged particles. The law is: F=kq1q2r2whereq1andq2represent two charges separated by a distance r and k is a constant approximately equal to 9×109Nm2C−2. (a) Find the work required to bring together a 0.8nCnegative charge and a 0.6nCnegative charge from a distance of 200 mm (0.2 m) to a distance of 100 mm (0.1 m).Note:1nC= 10−9CwhereCstands for coulombs and nC stands for...
[4 pts.] By what factor must the intensity of a sound be increased to raise its...
[4 pts.] By what factor must the intensity of a sound be increased to raise its loudness (i.e., its sound level) from 10 dB to 20 dB? From 20 dB to 30 dB? 0 dB to 10 dB?(b) [10 pts.] How much closer to a sound source must you stand to make the intensity your ear receives 20 times bigger? To increase its loudness by 20 dB?
Photographers often use multiple light sources to control shadows or to illuminate their subject in an...
Photographers often use multiple light sources to control shadows or to illuminate their subject in an artistic manner. The illumination from a light source is inversely proportional to the square of the distance from the source and proportional to the intensity of the source. Two light sources, L1 and L2, are 10m apart, and have intensity of 4 units and 1 unit, respectively. The illumination at any point P is the sum of the illuminations from the two sources. Hint:...
You have an experimental setup consisting of a light source, three polarization screens, and a light...
You have an experimental setup consisting of a light source, three polarization screens, and a light sensor. Logger pro records the intensity [lux] of the light incident to the light sensor vs. the angle of the polarization screen. The initial intensity of your light source (no polarizing screens) is  = 784 lux. Assume zero ambient light during the entire experiment. a) (4 pts) If you place ONE polarizing screen between the light source and light sensor, what is the expected intensity...
The sound intensity 30 meters from a jet engine is 10 Watts per square meter. What...
The sound intensity 30 meters from a jet engine is 10 Watts per square meter. What is the sound intensity level, in decibels, at the specified location a = 10 log [ I / IO ] IO = 10-12 W/m2 I = 10 Watts a = 130 dB B. Without hearing protection, repeated exposure to sounds at or above 80 decibels can cause hearing loss. What minimum distance from the source (the jet engine) would a listener have to keep...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT