Can any of you physics guru's help to get me started on my assignment question as I am struggling with it?

Question

A camera has a single convex lens of focal length of 50.0mm. The lens may be moved relative to the film plane such that the lens film distance may be varied between 50.0mm and 70.0mm. Find the range of object distances for which the camera may be adjusted to give a sharp image on the film. For the nearest possible object distance, find the linear magnification? many thanks in advance.

Answer 1

You probably want to use the so called lens makers equation for this.

If you are comfortable working with a sign convention, then it is 1/z' = 1/z + 1/f   where z' = distance to film (positive), f = positive (for convex lens), z = distance to object (negative).

Or, the common way it is stated:

1/f = 1/z' - 1/z   where all #s are positive now.

Anyway, here you're given z' can vary between 50 and 70, and f = 50.  Just solve above for z, in terms of f and z'.  Now, convince yourself that there are no local maximum or minimums (in this case .. there are not) -- so you just need to worry about the endpoints.  With the image plane at 50 mm (the rear focal point of the lens), you are focused at infinity.

For magnification -- recall how to draw an image through a lens.  Given an object at distance z, height h, you draw a line parallel to the optical axis until it hits the lens, then that goes through the rear focal point, and hits the image plane.  Also, draw a line from the object point, through the center of the lens, and keeps going straight.  These two rays intersect at distance z', and height h'.

Now, you've got two similar triangles -- they both have the ray through the center of the lens as the hypotenuse -- the one on the left has other sides h and z, and the one on the right has sides h' and z'.  Since these triangles have the same angles, they are similar -- so h'/h = z'/z.  The magnification is defined by h'/h -- so once you have solved for the minimum z above (and you know the corresponding z'), then m = z'/z.  Note that technically, the magnification is negative in the situation above (image is upside down).

Answer 2

For a focal length of 50mm the most distant object that can be focussed is at infinity ( a parallel beam of light will converge to the focal point).

You can prove this by using the formula :

1/f = 1/u + 1/v.

Now the closest object that can be focussed is when the lens is out at its farthest i.e 70mm = image distance (v)

so using the above formaula

1/50 = 1/u + 1/70

so 1/u = 1/50 - 1/70 which gives u = 175mm ( approx)

the magnification at this point is v/u = 70/175 = 0.4