Magnification and Power of a Lens Video Lecture - Class 10

magnification in lenses you know that
convex lenses are mainly used for
getting the enlarged images of the
objects these lenses are used by jewel
makers dentists etc concave lenses are
used for getting the diminished image of
a large area these lenses are used in
peepholes of doors while designing the
lens we need to know what is the size
and position of the image is produced by
it in this session we will learn about
magnification and the factors affecting
magnification in spherical lens in order
to understand magnification we need to
first understand a few definitions
object distance u the distance of the
object from the optical centre is called
the object distance image distance V the
distance of the image from the optical
centre is called the image distance
focal length F the distance of the
principal focus from the optical centre
is called the focal level now let's see
the formula for image distance and
height the formula which gives the
relationship between the object distance
u the image distance V and the focal
length F of a lens is called the lens
formula 1 by V minus 1 by U equals 1 by
F magnification is given by the ratio of
the height of the image to the height of
the object it is expressed by the letter
M M equals H dash by H magnification is
also expressed by the ratio of image
distance to that of object distance M
equals V by u
to understand the image formation in
lenses the following sign conventions
are used the object is always placed to
the left of the lens that is light from
the object falls on the lens from the
left-hand side all distances parallel to
the principal axis are measure from the
optical center of the lens all distances
measure to the right of the optical
centre are taken as positive of the
distances measured to the left are taken
as negative distances measured
perpendicular to and above the principal
axis are taken as positive distances
measured perpendicular to and below the
principal axis are taken as negative
focal length of a convex lens is taken
as positive focal length of a concave
lens is taken as negative let us take an
object of height 2 centimeter which is
placed perpendicular to the principal
axis 15 centimeters away from a
converging convex lens of focal length
10 centimeters as per the formula 1 by f
equals 1 by u minus 1 by V V equals 30
centimeters the size of the image formed
can be determined by the magnification
formula which is M equals H dash by H
equals V by u H dash equals H into V by
u height of the image H dash is H dash
equals minus 4 centimeters M equals V by
u equals minus 2 centimeters the
negative sign shows that the image is
inverted and real the image is 4
centimeters tall which is formed at a
distance of 30 centimeters on the other
side of the lens the image is 2 times
enlarged
to understand the image formation in
length
let us now keep focal length a constant
and vary the object distance used let's
see what will happen to the image
distance P magnification M and the
height of the object takes - now we will
keep object distance U of constant and
change the focal length and see the
subsequent changes to the image distance
magnification and the height of the
object now let us take a concave lens
whose focal length is 15 centimeters at
what distance should the object from the
lens be placed so that it forms an image
at 10 centimeters from the lens a
concave lens always forms a virtual
erect image on the same side of the
object as per the lens formula 1 by V
minus 1 by U equals 1 by F u equals
minus 30 centimeters
the object distance is 30 centimeters
the size of the image formed can be
determined by the magnification formula
which is M equals V by u M equals plus
0.33 the positive sign shows that the
image is erect and virtual the image is
one third of the size of the object
let's now keep focal length a constant
and vary the object distance U let's see
what will happen to the image distance V
magnification M and the height of the
object H - now we will keep object this
new view of constant and change the
focal length and see the subsequent
changes to the image distance
magnification and the height of the
object
from this demonstration we've understood
the following formula lens formula 1 by
V minus 1 by U equals 1 by F
magnification in a lens M equals H - by
H or V by u power of a lens you must
have used the zoom option in a camera
look at these images here we can see a
picture captured by a camera and another
one captured after zooming in the
ability to zoom in and capture good
images depends on the power of lens we
also hear the words power of lens when
we speak of spectacles people wear what
does this mean the ability of a lens to
converge or to diverged the rays of
light falling on it is depending on the
power of a lens
hence the degree of convergence or
divergence of light rays achieved by a
lens is expressed in terms of its power
power of lens is defined as the
reciprocal of its focal length if F is
the focal length of the lens then the
power P equals 1 by F the power of a
lens is dependent on the focal length of
a lens the SI unit of power is diopter
the focal length is expressed in meters
that power is expressed in diopter it is
represented by the letter d1 diopter is
the power of a lens whose focal length
is 1 meter the power of a convex lens is
positive and that of a concave lens is
negative to increase the magnification
or sharpness of the images in optical
instruments lenses are combined the net
power feed of the lenses placed in
contact is given by the algebraic sum of
the individual powers P 1 P 2 P 3
P equals P 1 plus P 2 plus P 3 plus so
on
some examples are cameras microscopes
and telescopes
let's take two convex lenses l1 and l2
of varying focal lengths when a beam of
light is passed through the lens l1 the
Rays get converged at a point let's
measure this distance we can see that
the light rays converge at a distance of
10 centimeters
hence the focal length of this lens is
10 centimeters applying the formula the
power is 10 D now
let's pass the beam of light beam l2 and
measure the distance to the point at
which the light rays converge the
distance is 20 centimeters hence the
power of this lens is 5 D from this
demonstration we've understood the
following formula P equals 1 by Fe you
have now understood the properties image
formation and magnification in lenses
you have also seen some uses of lenses
can you list more real-life application
of lenses
you