
No matter how far you stand from a mirror, your image always appears erect. The mirror is likely to be:
(A) Plane
(B) Concave
(C) Both A and B
(D) None
Answer
139.8k+ views
Hint You can clearly see that to find the correct answer, all you need to do is examine where and how the image is formed when the object is put at various different positions from the mirror, for both the mirror, concave and plane mirrors. Once you do that, tick the most suitable option, i.e. the mirror which always creates an erect image of an object, no matter where it is placed from the mirror.
Complete step by step answer
To solve the question, we will proceed with the same approach as explained in the hint section of the solution to the question. First let us observe the image formation through reflection from a plane mirror:
We can quite clearly observe that the radius of curvature of a plane mirror is nothing but infinity, since there is no curvature in a plane mirror, we can write it as:
$r = \infty $
If the radius of curvature is infinity, this most certainly means that the focal length is infinity as well:
$
f = \dfrac{r}{2} = \dfrac{\infty }{2} \\
\Rightarrow f = \infty \\
$
Now, let us assume that the object is placed at a distance $u$ from the mirror.
Let us assume that the image of the object is formed at a distance $v$ from the mirror.
Now, if we apply the mirror formula, we get:
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
If we substitute the value of focal length as $f = \infty $ , we get:
$0 = \dfrac{1}{v} - \dfrac{1}{u}$
The negative sign is used since the image distance is $u$ but on the negative coordinate side.
Solving this, we get:
$v = u$
Which means that the image is formed at the exact same distance.
Let us check the magnification as:
$m = \dfrac{{ - v}}{u}$
Where $m$ is magnification
If we substitute the value of $v = u$ and $u = - u$ , we get:
$m = \dfrac{{ - u}}{{ - u}} = 1$
A positive value of magnification means that the image formed is erect. Hence, Plane mirror always forms an erect image, no matter where the object is placed.
We can see in the table that the only cases when the concave mirror makes an erect image is when the object lies at the focus or between the focus and the pole. In all the other cases, the image is formed inverted and not erect, hence, we can confidently say that the concave mirror is not the answer.
Now, if we check the options, we will see that the only correct one is the option (A) as only a plane mirror creates an erect image from the given options.
Note It is to be noted that even the convex mirror makes an erect image in every case, no matter where the object is placed from the mirror. Since convex mirror was not given in the option, we only ticked the plane mirror’s option as the correct answer.
Complete step by step answer
To solve the question, we will proceed with the same approach as explained in the hint section of the solution to the question. First let us observe the image formation through reflection from a plane mirror:
We can quite clearly observe that the radius of curvature of a plane mirror is nothing but infinity, since there is no curvature in a plane mirror, we can write it as:
$r = \infty $
If the radius of curvature is infinity, this most certainly means that the focal length is infinity as well:
$
f = \dfrac{r}{2} = \dfrac{\infty }{2} \\
\Rightarrow f = \infty \\
$
Now, let us assume that the object is placed at a distance $u$ from the mirror.
Let us assume that the image of the object is formed at a distance $v$ from the mirror.
Now, if we apply the mirror formula, we get:
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
If we substitute the value of focal length as $f = \infty $ , we get:
$0 = \dfrac{1}{v} - \dfrac{1}{u}$
The negative sign is used since the image distance is $u$ but on the negative coordinate side.
Solving this, we get:
$v = u$
Which means that the image is formed at the exact same distance.
Let us check the magnification as:
$m = \dfrac{{ - v}}{u}$
Where $m$ is magnification
If we substitute the value of $v = u$ and $u = - u$ , we get:
$m = \dfrac{{ - u}}{{ - u}} = 1$
A positive value of magnification means that the image formed is erect. Hence, Plane mirror always forms an erect image, no matter where the object is placed.
We can see in the table that the only cases when the concave mirror makes an erect image is when the object lies at the focus or between the focus and the pole. In all the other cases, the image is formed inverted and not erect, hence, we can confidently say that the concave mirror is not the answer.
Now, if we check the options, we will see that the only correct one is the option (A) as only a plane mirror creates an erect image from the given options.
Note It is to be noted that even the convex mirror makes an erect image in every case, no matter where the object is placed from the mirror. Since convex mirror was not given in the option, we only ticked the plane mirror’s option as the correct answer.
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