What are the characteristics of an image formed by a plane mirror?

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The virtual image produced by a plane mirror appears to be the same distance away from the mirror as the object itself.

A plane mirror produces a top-to-bottom reversal and a left-to-right reversal.

The height of an image as seen in a plane mirror is the same as the actual height of the object.

Both B and C.

answer is D.

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Detailed Solution

Concept: In order to solve the challenges and ascertain the type of the image, we will draw the ray diagrams. We'll determine the image's position, orientation, and relative shape. We will look for similar triangles and compare the sides by applying the proper sign conventions for the sides in order to determine the concave mirror formula.
The characteristics of an image formed by a plane mirror are mentioned hereunder:
(1) There is no difference between image and object distance.
(2) The item and image are of the same size.
(3) The resulting image is upright and virtual.
(4) Laterally inverted images.
(5) A shadowy image appears behind the mirror.
Figure displays the three-ray ray diagram. It shows the representation A′B′ (in this instance actual) of an object AB made up of a concave mirror. Therefore, point A′ is the image point of point A if every ray coming from point A that hits the concave mirror after reflection travels through point A′.
Question Image

The mirror equation, or the relationship between object distance (u) , image distance (v) , and focal length, is now derived ( f ).
The two right-angled triangles A′B′F and MPF are comparable in Figure. (MP is a straight line perpendicular to CP for paraxial rays.) Therefore,

\frac{B' A'}{PM}

\frac{B' F}{FP}

we can also write PM=AB

\frac{B' A'}{AB}

\frac{B' F}{FP}

…………………………. Equation (1)
The right-angled triangles A′B′P and ABP are comparable because ABP and A’B’P are equal.

\frac{B' A'}{BA}

\frac{B' P}{BP}

……………………..….. Equation (2)
As on comparison of two Equations (1) and (2)

\frac{B' F}{FP}

\frac{B' P - FP}{FP}

\frac{B' P}{BP}

……………… Equation (3)
A relation involving the magnitude of distances is seen in equation (3). Now let's use the sign convention. Light moves from the item to the mirror MPN, as we can see. Thus, this is viewed as moving in a favourable direction.
From the pole P, we must move in the opposite direction of the incident light in order to reach the object AB, image A′B′, and focus F. As a result, all three will be in bad shape. Thus,
B'P = -v, FP = -f , BP = -u
On putting the values we get