The magnification of the plane mirror is always...............
A. More than \[1\]
B. \[1\]
C. Less than \[1\]
D. Zero
Answer
513k+ views
Hint: To have a vague picture of the kind of images that the convex, concave, and plane mirrors are producing. If we can deduce the same for the plane mirror, we can get our answer for the question easily. We should also know its nature of being formed too.
Complete step by step solution:
The plane mirror can be called a mirror that has a flat or a planar reflective surface. The image that is formed here by the plane will always be virtual and the lateral inversion occurs during this process. This lateral inversion is experienced in our daily lives in front of the mirror, as the mirror shows the left hand when we hold up the right.
The magnification that is being produced by the mirror is given by the relation
$m = - \dfrac{{Distance{\text{ }}of{\text{ }}the{\text{ image}}}}{{Distance{\text{ }}of{\text{ }}the{\text{ }}object}} = - \dfrac{v}{u}$
In the case of the plane mirror, we can say that the distance of both the image and the object are equal and the object distance is said to be negative. Therefore it yields the magnification of $ + 1$ and $M = 1$also signifies that the size of both the image and the object are also equal and the resultant image is erect.
So the answer is option B.
Note:
The distances are measured from the pole of the mirror. According to the conventional method, the negative sign specifies the distance measured in the direction opposite to the incident ray while the positive sign specifies the distance measured in the direction of the incident ray. The distance below the axis is negative while the distance above is positive and the radius of curvature and focal length is taken as infinity.
Complete step by step solution:
The plane mirror can be called a mirror that has a flat or a planar reflective surface. The image that is formed here by the plane will always be virtual and the lateral inversion occurs during this process. This lateral inversion is experienced in our daily lives in front of the mirror, as the mirror shows the left hand when we hold up the right.
The magnification that is being produced by the mirror is given by the relation
$m = - \dfrac{{Distance{\text{ }}of{\text{ }}the{\text{ image}}}}{{Distance{\text{ }}of{\text{ }}the{\text{ }}object}} = - \dfrac{v}{u}$
In the case of the plane mirror, we can say that the distance of both the image and the object are equal and the object distance is said to be negative. Therefore it yields the magnification of $ + 1$ and $M = 1$also signifies that the size of both the image and the object are also equal and the resultant image is erect.
So the answer is option B.
Note:
The distances are measured from the pole of the mirror. According to the conventional method, the negative sign specifies the distance measured in the direction opposite to the incident ray while the positive sign specifies the distance measured in the direction of the incident ray. The distance below the axis is negative while the distance above is positive and the radius of curvature and focal length is taken as infinity.
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