Solveeit Logo
Login

Question

Question: Magnification, \(m=\) ………………. A. \(\dfrac{v}{u}\) B. \(\dfrac{u}{v}\) C. \(\dfrac{{{h}_{o}}}...

Magnification, m=m= ……………….
A. vu\dfrac{v}{u}
B. uv\dfrac{u}{v}
C. hohi\dfrac{{{h}_{o}}}{{{h}_{i}}}
D. hiho\dfrac{{{h}_{i}}}{{{h}_{o}}}

Explanation

Solution

Try to remember the definition of magnification in order to solve the question. Here, sign conventions play an important role. The object distance, image distance, object height and image height may have different signs according to their positioning that may affect the value of magnification.

Complete step by step solution:
In optics, an object is considered a body which radiates light rays. The light may be emitted by the object itself or reflected from it. There are two types of object: point object and extended object. When the light rays emitted from the object strikes a mirror or lens, it gets reflected and appears to converge or diverge from a position different from the position of the original object and it appears that the object is there. This apparent object which is observed by an observer is called an image of the object.
Here, Cartesian sign convention is used to define the positions of objects and images. The distances measured in the direction of the light is considered to be positive and opposite to the direction of light as negative. The principal axis acts as a reference to measure the object and image height. Heights measured above the principal axis are positive and below it are negative.
Magnification refers to the enlargement of the object size with the help of different lenses and mirrors. The definition of magnification is defined as the ratio of the image height to object height. Also, it is the ratio of the image distance to object distance.
m=hiho=vum=\dfrac{{{h}_{i}}}{{{h}_{o}}}=\dfrac{v}{u}
Where hi={{h}_{i}}= Image height
ho={{h}_{o}}= object height
v=v= Image distance
u=u= object distance
In a mirror, the image is located at the same distance as the object but in the opposite direction. Also, the image and object height are the same. Therefore the image is virtual and erect and its magnification is +1+1 . An image is virtual as no light rays actually come from it.
Therefore options A and D are correct.

Note:
If the magnitude of the magnification is positive, the image is virtual and erect and if the magnitude of the magnification is negative, the image is real and inverted. The process of magnification is used to manufacture different instruments like magnifying glass, telescope etc.