An object is placed between a magnifying lens and its focal point. (a) Show that
ID: 3898141 • Letter: A
Question
An object is placed between a magnifying lens and its focal point. (a) Show that if the object is between the lens and half the focal distance (i.e. p < 1?2 f ), then the image
will be between the lens and the focal point (i.e. q < f ). Will the image be real or virtual?
Inverted or upright?
(b) Suppose the object is placed a distance 1?2 f from the lens. What will the magnification be? What
will the angular magnification be, assuming the near point to be at N?
(c) Where should the object be placed if the image is to be a distance 2f from the lens? What would
the magnification be in this case? What is the angular magnification?
(d) Now suppose the object was put just inside the focal distance from the lens (set p = xf where x is
smaller than 1, but very close to 1). What would be the distance of the image from the lens? What would be the magnification? What would be the angular magnification?
Explanation / Answer
a) Magnifying lenses are converging lenses. When an object is place between the focal point and the lens, the image will be virtual and upright. You can see this for yourself or draw ray-diagrams.
Now to determine whether q < f you need to either draw the ray diagram of such a situation or show the math. Do so for p = f/4
b) 1/f = 1/p + 1/q where f is the focal length, p is the object distance and q is the image distance.
1/f = 1/(f/2)+1/q
1/f-2/f=1/q
(1-2)/f=1/q
-1/f=1/q
-q = f
q=-f (the image is virtual because q is -)
Magnification M = -q/p = f/(f/2) = 2 (the image is upright because M is positive)
c) q = 2f, d = ?
1/f = 1/d + 1/2f
1/f - 1/2f = 1/d
2-1/2f = 1/d
d = 2f
and M = -d/q = -1
d) d=xf where x->1, you can use limits if you know calculus or use x = 0.99
1/f = 1/0.99f + 1/q
1/f - 1/0.99f = 1/q
(0.99-1)/0.99f = 1/q
q= -0.99f/0.01
q = -99 f
The image is virtual because q is negative.
M = -q/d = 99f/0.99f = 100