Question

In case of a virtual and erect image, the magnification of a mirror is:
A. Positive
B. Negative.
C. Unity.
D. Infinity

Answer 1

Hint: A image is formed by a mirror when the two the reflected rays of the same object meets. If the two reflected rays actually meet then the image is called a real image. And if the two reflected rays are diverse after reflection and will never meet and we have to extend them in backward to see where the two reflected rays appear to meet then the meeting point is called the virtual image. The image may be real or virtual, erect(upright) or inverted. The magnification of a mirror is given by the ratio of image height to object height. Use sign convention to see whether the magnification is positive or negative or unity or infinity.

Formulas used:
Magnification of a mirror is
\[m=\dfrac{h'}{h}\]
Where $h'$ is the height of the image and $h$ is the height of the object.

Complete step by step answer:
A virtual image is the image of an object when the two reflected rays are diverse after reflection and will never meet and we have to extend them backwards to see where the two reflected rays appear to meet is called the virtual image.
An image may be real or virtual and it may be erect or inverted.
According to the sign convention the upward distance from the principal axis is taken to be positive and the downward distance from the principal axis is taken to be negative.
The principal axis is the line drawn perpendicular to a mirror at the midpoint of the mirror.
And according to sigh convention the object height is always taken above the principal axis. So the object height is always positive. As the magnification of a mirror is the ratio of image height to object height, and we see that the object height is always taken to be positive so the magnification will only depend upon how the image is oriented. If the image is erect then the image height will be taken as positive according to sign convention so the magnification of a virtual erect image is always positive.
So the correct option is A. Positive

Note:
In every ray optics problem the sign convention should be used in the formula. The sign convention is like the co-ordinate axis. If you go right from the lens or mirror the distance will be taken as positive and if you go left from the lens or mirror the distance will be taken as negative. Similarly moving top is positive and moving bottom or down will be negative.