Magnification is the size of the image relative to the size of the object. The magnification can be linear, longitudinal or angular. Linear magnification is the ratio of the image length to the object length measured in the plane that is perpendicular to the optical axis. In longitudinal magnification, the image length is measured along the optical axis. The angular magnification is measured by taking the tangent of the angles made by the object and the corresponding images when it is measured from a specific point in the instrument, like binoculars and magnifiers.
Longitudinal Magnification of Spherical Mirror
Magnification is the enlargement of the apparent size of the object. Longitudinal magnification is when the object is placed parallel to the principal axis. It is defined as the ratio of the length of the image to the length of the object.
Longitudinal magnification = li/lo
li is the length of the image
lo is the length of the object
Case 1: If the object length is not very small
If the object length is not very small, then the longitudinal magnification can be determined by using the formula,
Longitudinal magnification, ML = (v2 – v1)/(u2 – u1).
(v2 – v1) = difference of the extreme positions of the image
(u2 – u1) = difference of the extreme positions of the object
Case 2: If the object length is very small
In order to determine the longitudinal magnification formula for very small objects, we must consider the mirror formula.
(1/v) + (1/u) = (1/f)
The above equation can also be written as
v-1 + u-1 = (1/f)
Let us differentiate the equation w.r.t to u
(-1)v-2 (dv/du) + (-1)u-2 = 0
Therefore, we get the formula for longitudinal magnification as
(dv/du) = – v2/u2
Here,
dv = length of the image
du = length of the object
The negative sign indicates that the image will be inverted
This formula is applied when the length of the image is very small compared to the distance of the object from the pole.
Solved Examples
1. Consider a concave mirror of a focal length of 20 cm. A rod of length 20 cm is placed along the principal axis at a distance of 40 cm from the pole of the concave mirror. Find the longitudinal magnification.
Solution:
Given,
f = -20 cm
From the image, we get
u1 = – 40 cm
u2 = – 60 cm
v1 = u1f/(u1-f)
= (-40)(-20)/(-40-(-20))
= 800/(-20)
= – 40 cm
v2 = u2f/(u2-f)
= (-60)(-20)/(-60 -(-20))
= 1200/-40
= -30 cm
Therefore, the length of the image = v2 – v1
= – 40 – (-30) = – 10 cm
Therefore, the longitudinal magnification (ML) is
ML = (v2 – v1)/(u2 – u1)
= 10/20
= – ½
The negative sign indicates that the image is inverted.
2. Consider a convex mirror of a focal length of 20 cm. A rod of length 10 cm is kept at a distance of 20 cm from the pole of the mirror. Find the longitudinal magnification.
Solution:
Given,
f = 20 cm
From the image, we know
u1 = – 20 cm
u2 = – 30 cm
v1 = u1f/(u1-f)
= (-20)(20)/(-20-20))
= -400/(-40)
= 10 cm
v2 = u2f/(u2-f)
= (-30)(20)/(-30 – 20)
= -600/-50
= 12 cm
Therefore, the length of the image = v2 – v1
= 12- 10 = 2 cm
Therefore, the longitudinal magnification (ML) is
ML = (v2 – v1)/(u2 – u1)
= 2/10
= – ⅕
The image is inverted, so we should put the negative sign.
Related Video
Frequently Asked Questions on Longitudinal Magnification
Explain what longitudinal magnification is.
The factor by which an image increases in size as measured along the optical axis is known as longitudinal magnification.
Can the longitudinal magnification be negative?
Yes, the longitudinal magnification can be negative when the object is inverted.
What is linear magnification?
The ratio of image size to object size is known as linear magnification or transverse magnification.
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