0
$\begingroup$

I actually want to know all the possible cases of image formation by a convex mirror[using a virtual object] similar to something like this table which shows the nature of images formed by a concave mirror for certain specific points[for a real object].image formation by concave mirror

I cannot understand when the image will be real/virtual as well as erect/inverted. Can you specify the nature of these 5 points by taking these points behind the mirror instead of in front obviously as the object is virtual in case of convex.

$\endgroup$

2 Answers 2

1
$\begingroup$

A convex mirror always makes virtual pictures, wich are always the same orientation the object has. So only the distance of these pictures varies.

$\endgroup$
2
  • $\begingroup$ This is not correct. See, for example, en.wikipedia.org/wiki/Cassegrain_reflector $\endgroup$
    – DJohnM
    Commented Mar 3, 2023 at 16:42
  • $\begingroup$ @DJohnM The convex secondary mirror in the Cassegrainian telescope corrects aberrations of the concave primary but it does not, by itself, form the image. $\endgroup$
    – John Doty
    Commented Mar 3, 2023 at 17:30
1
$\begingroup$

To determine whether the object is inverted, follow the rays from the top of the object (from $A$) to the top of the image ($A'$), and from the bottom of the object ($B$) to the bottom of the image ($B'$). If $A'$ is above $B'$, the object is erect, otherwise it is inverted.

The image is real if it is formed by the rays themselves after reflection from the mirror. It is virtual if it is formed by extrapolating the reflected rays into the other side of mirror. This is shown by the dashed lines in case (vi) of the graphic you uploaded. In all other cases the images are real because they are formed by the actual rays.

How do we know whether to extrapolate or not? We always go for the side where the rays converge. Notice that picture always shows two rays starting from $A$. If their reflections converge, this defines $A'$ and represents a real image. If the reflections diverge, then their extrapolations into the mirror converge and the image is virtual.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.