Is the Image Formed by a Concave Lens Real or Virtual?

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SUMMARY

The discussion centers on the nature of images formed by concave lenses, specifically addressing whether the image is real or virtual. Participants conclude that the image formed by a concave lens is virtual, erect, and not magnified, contradicting the initial assumption that it could be real. The mathematical approach used to derive the image distance and magnification is deemed unnecessary, as the fundamental principles of ray behavior through diverging lenses clarify that no image is formed. The question posed is considered flawed due to its misleading premise.

PREREQUISITES
  • Understanding of lens types, specifically concave and convex lenses
  • Familiarity with ray diagrams and principal rays for diverging lenses
  • Basic knowledge of optics, including image formation and magnification
  • Ability to interpret focal lengths and image distances in lens equations
NEXT STEPS
  • Study the principles of ray optics, focusing on diverging lenses
  • Learn about the mathematical derivation of lens formulas, including the lens maker's equation
  • Explore practical applications of concave lenses in optical devices
  • Investigate common misconceptions regarding image formation in optics
USEFUL FOR

Students of physics, educators teaching optics, and anyone interested in understanding the behavior of light through lenses will benefit from this discussion.

Amith2006
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# 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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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.
 
Chi Meson said:
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.
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?
 
Please respond.
 

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