
Are real images always inverted?
Answer
487.8k+ views
Hint:Two types of images are generally formed by the mirror. They are named real images and virtual images. The real images are those which are formed in front of the mirror. Whereas, the virtual image is formed in the back of the mirror. The formula of the magnification is needed to answer the question. In this formula, the sign convention is very essential because from this we can know why the real images are inverted.
Formula used:
The magnification for both the convex mirror and concave mirror is, $m = - \dfrac{v}{u}$
where, $v$ is the image-distance and $u$ is the object-distance from the mirror.
Complete step by step solution:
Let us take a real object which is at distance $u$ from the mirror and the image is formed at the distance $v$.
The real image is formed in front of the mirror. Hence, we have to take both the distances in the positive sign (also, one can take them as negative signs).
Now, the pattern of the image ( i.e. either it will be inverted or the same as the object) will depend upon the sign of the magnification.
The magnification for both the convex mirror and concave mirror is, $m = - \dfrac{v}{u}$
Since the signs are the same for both distances, the magnification will be found as negative. And, this negative sign implies that the image will be inverted.
On the other hand, it is noticed that the real images are always formed below the principle axis. This is also a reason for the inversion as the image is the copy of an object.
Hence, we can say the real images are always inverted.
Note:When the mirror forms a virtual image of a real object, the image is generally erect. Since the virtual image is formed on the back side of the mirror and the object is always in the front side of the mirror, their signs are different from each other.
Hence from the formula of the magnification, $m = - \dfrac{v}{u}$
We get, the magnification becomes positive in sign. Therefore the virtual images are always erect or the same-sided as the object and are formed above the principle axis.
Formula used:
The magnification for both the convex mirror and concave mirror is, $m = - \dfrac{v}{u}$
where, $v$ is the image-distance and $u$ is the object-distance from the mirror.
Complete step by step solution:
Let us take a real object which is at distance $u$ from the mirror and the image is formed at the distance $v$.
The real image is formed in front of the mirror. Hence, we have to take both the distances in the positive sign (also, one can take them as negative signs).
Now, the pattern of the image ( i.e. either it will be inverted or the same as the object) will depend upon the sign of the magnification.
The magnification for both the convex mirror and concave mirror is, $m = - \dfrac{v}{u}$
Since the signs are the same for both distances, the magnification will be found as negative. And, this negative sign implies that the image will be inverted.
On the other hand, it is noticed that the real images are always formed below the principle axis. This is also a reason for the inversion as the image is the copy of an object.
Hence, we can say the real images are always inverted.
Note:When the mirror forms a virtual image of a real object, the image is generally erect. Since the virtual image is formed on the back side of the mirror and the object is always in the front side of the mirror, their signs are different from each other.
Hence from the formula of the magnification, $m = - \dfrac{v}{u}$
We get, the magnification becomes positive in sign. Therefore the virtual images are always erect or the same-sided as the object and are formed above the principle axis.
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