Please help me with physics!!?? please??

Question

An arrow 2.00 cm long is located 75.0 cm from a lens, which has a focal length f = 34.1 cm.

(a) If the arrow is perpendicular to the principal axis of the lens, as in the Figure A, what is its lateral magnification, defined as hi / ho? (Include the sign of your answer.)

(b) Suppose, instead, that the arrow lies along the principal axis, extending from 74.0 cm to 76.0 cm from the lens, as indicated in Figure B. What is the longitudinal magnification of the arrow, defined as Li / Lo? (Hint: Use the thin-lens equation to locate the image of each end of the arrow. Include the sign of your answer.)

diagram: http://www.webassign.net/walker/26-56.gif

Answer 1

This will help you out very very much.

Hi/Ho = Si/So

Answer 2

I am using coordinate sign convention. Let me know if you want some other sign convention.

a) Object distance u = -75 cm

Focal length f = 34.1 cm

Let v = image distance

1/v - 1/u = 1/f

Or, 1/v = 1/f + 1/u

Or, 1/v = (f + u)/uf

Or, v = uf/(f + u)------------------(1)

Magnification = v/u = f/(f + u)

= 34.1/(34.1 - 75)

= 34.1/(-40.9)

= -0.83

Ans: -0.83. Negative sign means that inverted image will be formed.

b) For left end:-

Object distance u1 = -76 cm

Let v1 = image distance

From (1)

v1 = u1f/(f + u1) = -76 * 34.1/(34.1 - 76) = -76 * 34.1/(-41.9) = 61.85 cm

For right end:-

Object distance u2 = -74 cm

Let v2 = image distance

From (1)

v2 = u2f/(f + u1) = -74 * 34.1/(34.1 - 74) = -74 * 34.1/(-39.9) = 63.24 cm

Length of image = v2 - v = 63.24 - 61.85 = 1.39 cm

Length of object = u2 - u1 = -74 - (-76) = 76 - 74 = 2 cm

Li/L0 = 1.39/2 = 0.7 (approx.)

Answer 3

(a)Assume Length from object to the lens is p = 75 cm

length from lens to image q = ? cm

focal length f = 34.1 cm

magnification M = ?

(1/p)+(1/q) = 1/f

1/q = (1/f)-(1/p)

= 1/34.1 - 1/75

=62.5 cm

M = hi/ho = -q/p

= -(62.5/75)

= - 0.83

Please check calculation and equation from the book

Part b is not my physics book, sorry. If I find it, then I will post. But good luck to you first.

And, the above solution is my own, but you can also find different way of solving it here

http://www.physicsforums.com/showthread....

You might be able to post your question there, or it may have already been answered.