Real Image and Virtual Image
Real image is that image which can be obtained on a screen. It is formed by the actual intersection of the refracted rays. It is usually formed by a convex lens.
Virtual image is that image which cannot be obtained on a screen. It is formed by the intersection of the rays. It is usually formed by a concave lens.
Differences between Real Image and Virtual Image
The major differences between real image and virtual image are:
S. No. |
Real Image |
S. No. |
Virtual Image |
1. |
It is formed at the point where the refracted rays meet. |
1. |
It is formed at the point where the refracted rays appear to meet. |
2. |
It is always inverted. |
2. |
It is always erect. |
3. |
The image is usually formed either on another side or behind the lens. |
3. |
The image is always formed on the same side of the object in the lens. |
4. |
Its size depends on the distance of the object from the optical center of the lens. |
4. |
Its size is larger in the convex lens and smaller in the concave lens. |
5. |
The image can be obtained on a screen. |
5. |
The image cannot be obtained on the screen. |
Magnification (m)
The size of the image obtained by the lens depends upon the distance of the object from the lens. If the object is placed near the lens, the image is magnified and if the object is placed far from the lens, the image is diminished. Thus, we can define magnification of a lens as the ratio of a height of the image to the height of the object.
Mathematically,
Magnification = \(\frac {height\:of\:image\:(I)}{height\:of\:object\:(O)}\)
\(\therefore\) m = \(\frac IO\)
Explanation:
In the above figure, two triangles \(\triangle\)OCD and \(\triangle\)OAB formed are similar, since all the angles of two triangles are equal. We have: \(\frac {CD}{AB}\) = \(\frac {OC}{OA}\)
According to the definition of magnification:
Magnification = \(\frac {CD}{AB}\)
Or, m = \(\frac {distance\:of\:image\:from\:lens\:(v)}{distance\:of\:object\:from\:lens\:(u)}\)
i.e. m = \(\frac vu\)
Therefore, magnification is also calculated by the ratio of image distance (v) to object distance (u).
Interpretation of Magnification
Hence, we can conclude that magnification shows how smaller or larger an image is than the object.
To prove: \(\frac IO\) = \(\frac vu\)
Let an object AB be placed on the principal axis of the convex lens and perpendicular to its principal axis beyond 2F. A ray BP is parallel to the principal axis passes through F after refraction through it and another ray BO passes undeviated through its optical center O. These two refracted rays PB’ and OB’ meet at B’. Hence, B’ is the real image of B, and A’B’ is the real image of AB.
In \(\triangle\)ABO and \(\triangle\)A’B’O, we have;
\(\therefore\) \(\triangle\) ABO = \(\triangle\) A’B’O are similar.
Hence,
\(\frac {A’B’}{AB}\) = \(\frac {OA’}{OA}\)
i.e. \(\frac {height\:of\:image}{height\:of\:object}\) = \(\frac {image\:distance}{object\:distance}\)
\(\therefore\) \(\frac IO\) = \(\frac vu\) _{Proved.}
Relation between Object distance, Image distance, and Focal length
If u, v and f represent object distance, image distance and focal length of a lens respectively, we can give the relation between them by a formula:
\(\frac 1f\) = \(\frac 1u\) + \(\frac 1v\)
It is said to be lens formula.
For our simplicity, we take the real distance as positive and virtual distance is taken as negative. Hence, the focal length of convex lens is taken as positive and the focal length of concave lens is taken as negative.
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