Solving Diverging Mirror Homework: Size & Location of Image

  • Thread starter Thread starter donotremember
  • Start date Start date
  • Tags Tags
    Mirror
AI Thread Summary
The discussion revolves around calculating the size and location of an image formed by a diverging mirror with a focal length of -60 cm. The initial attempt at solving the problem yielded an image distance (di) of -0.5454 m, while the textbook provided a different answer of -0.66 m. Participants noted that the object distance (do) should be positive, as it refers to a real object, which contradicts the textbook's solution. The confusion stems from the sign convention used in optics, particularly regarding virtual images. Ultimately, the consensus is that the object distance must be positive, affirming the correctness of the initial calculation.
donotremember
Messages
31
Reaction score
0

Homework Statement



A trucker sees the image of a car passing her truck in her diverging rear view mirror whose focal length is -60cm. If the car is 1.5m high and 6.0m away, what is the size and location of the image?

Homework Equations



1/f = 1/di + 1/do

The Attempt at a Solution



Finding the di

The solution I come up with is:

1/-0.6 = 1/di + 1/6


-1.66 - 0.166 = 1/di

di = -0.5454


The correct answer in my book is :

1/-0.6 = 1/di + 1/-6


-1.66 + 0.166 = 1/di

di = -0.66

Why is the distance to the object negative?

This does not agree with sign convention (as stated in my book)

"Distances of real objects and images are positive"
 
Physics news on Phys.org
Your answer is correct.
 
rl.bhat said:
Your answer is correct.

You mean the distance is not negative and the real answer is -0.5454?

That is strange because the textbook got it wrong and the lecture notes did too.
 
You got di negative because the image is virtual.
 
donotremember said:
You mean the distance is not negative and the real answer is -0.5454?

By this question i was referring to the 6m distance to the object.
 
This is weird, meaning the book's solution. Object distances are always real (and therefore positive) -- except that when the "object" is actually the image produced by some other mirror or lens, it could be negative. But that's not the case here, the object is an actual object, therefore do must be positive.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top