For magnification in spherical mirror object height is:
(A) Negative
(B) Positive
(C) For real images positive
(D) For virtual images negative
Answer
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Hint If we want to magnify an image of an object, we use a spherical mirror called a concave mirror. Next, we need to recall the conventions which we use for making ray diagrams involving concave mirrors. The object height is always taken positive if the object is in the positive y-direction.
Complete step by step answer
Spherical mirrors: it is a part of a spherical surface mirror. It is of two types- one is concave and another one is convex. The bulged out part is known as convex mirror whereas it is thinner from the center and thicker at the edges then it is known as concave mirror.
Magnification is defined as the ratio of image height to an object height. It is denoted by ‘m’.
$m = \dfrac{{{h_i}}}{{{h_o}}}$ Where ${h_{i\,}}$ -image height; ${h_o}$ - object height; m- magnification.
It is used to increase the height of the object.
Real image: when an image is formed on the screen.
Virtual image: when an image appears to form on the screen.
Image gets shorter and longer depending upon mirrors, but the object is always placed above the principal axis.
So, it will always be positive. Therefore, option B is correct.
Note
If object height is taken as negative then it should be placed below the principal axis, but that is against the well-established convention, so it is unwise to think in that direction. Hence option A is considered wrong.
Complete step by step answer
Spherical mirrors: it is a part of a spherical surface mirror. It is of two types- one is concave and another one is convex. The bulged out part is known as convex mirror whereas it is thinner from the center and thicker at the edges then it is known as concave mirror.
Magnification is defined as the ratio of image height to an object height. It is denoted by ‘m’.
$m = \dfrac{{{h_i}}}{{{h_o}}}$ Where ${h_{i\,}}$ -image height; ${h_o}$ - object height; m- magnification.
It is used to increase the height of the object.
Real image: when an image is formed on the screen.
Virtual image: when an image appears to form on the screen.
Image gets shorter and longer depending upon mirrors, but the object is always placed above the principal axis.
So, it will always be positive. Therefore, option B is correct.
Note
If object height is taken as negative then it should be placed below the principal axis, but that is against the well-established convention, so it is unwise to think in that direction. Hence option A is considered wrong.
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