# Homework Help: Simple lens question (camera)

1. May 22, 2010

### ZedCar

1. The problem statement, all variables and given/known data
Was wondering if someone could help...

2. Relevant equations
1/f = 1/u + 1/v
m = v/u

3. The attempt at a solution
Using:
1/f = 1/u + 1/v

1/50 = 1/8000 + 1/(v)
v = 50.3 mm

Then to calculate the linear magnification of the image, using m = v/u with same v and u values from above? (both in metres)

i.e.

m = 0.0503 / 8 = 0.0063

Though as the image gets flipped upside down it's probably:
m = -0.0503 / 8 = -0.0063

Thank you

2. May 22, 2010

### Redbelly98

Staff Emeritus
Looks good. In fact the correct equation for magnification does have the minus sign:

m = -v/u

3. May 22, 2010

### ZedCar

Thanks very much Redbelly98. I hadn't realised that!

4. May 24, 2010

### ZedCar

Is that definitely correct?

I looked the equation up in my CGP Physics Revision Guide and it states the equation as:

m = v/u

5. May 24, 2010

### silence98

Yes, it is m=-v/u.

6. May 24, 2010

### ZedCar

So substituting with the numbers from the initial post:
m = -v / u
m = -0.0503 / 8 = -0.0063

So the answer for m is definitely a -'ve answer and not +'ve ?

7. May 24, 2010

### Redbelly98

Staff Emeritus
Yes. The image is inverted, and m is negative.

8. May 25, 2010

### ZedCar

I had thought, from what I learned in class, that the -'ve sign indicated that the image was virtual.

I hadn't realised it was an indication of whether the image was inverted.

Which is why I thought the m would be +'ve since the camera image is real.

I haven't been studying diagrams for long, so it would appear that I have not quite got the concept correct yet.

9. May 25, 2010

### Redbelly98

Staff Emeritus
The fundamental definition of magnification is the ratio of the heights, himage/hobject, with an inverted image treated as a negative height. Many physics textbooks use a geometric argument, using similar triangles, to show that this is equivalent to -v/u. The -v/u formula is generally more useful, since it's more common to give information about the distance from the lens rather than the height.