# Image formed by concave lens

1. ### Amith2006

420
# A convergent beam is incident on a concave lens as shown in figure. Which of the following is not correct?
a) The image formed is real
b) The image formed is virtual
c) The image formed is erect
d) The image formed is magnified
I solved it in the following way:

Let f be the focal length of the concave lens and let v be the image distance. From the figure, object distance = -f(virtual object), Focal length = -f(focal length of concave lens is negative)
(1/ object distance) + (1/image distance) = 1/ Focal length
(1/-f) + (1/v) = 1/-f
Solving I get,
v = + infinity
Magnification = -v/u = (- infinity)/(-f)
= + infinity
Since Magnification is positive the image is erect. Since the Magnification is infinity the image is magnified. Since the image distance is positive the image is real. So the answer is (b). Is my argument right?

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2. ### Chi Meson

1,767
What a bizarre question.

First of all, you don't need the math. Look at the drawing and determine "what happens to rays that go through a diverging lens if those rays are heading toward the far focal point." This is one of the basic principal rays for diverging lenses.

The result is that both rays emerge parallel to the principle axis. Parallel rays do not intersect to form real or virtual images. There is no image here, and that is why you got the infinite image distance. The question is flawed but your work is impressive.

3. ### Amith2006

420
In the case of a convex lens, if an object is placed at its first principal focus, the light rays from the object after passing through the lens is rendered parallel. We then say that the a real, inverted and extremely magnified image is formed at infinity. Can't we say the same thing in the case of concave lens?

420