Class 12 Exam > Class 12 Videos > L36 : Numerical lens - Ray Optics, Physics, Class 12
L36 : Numerical lens - Ray Optics, Physics, Class 12 Video Lecture
FAQs on L36 : Numerical lens - Ray Optics, Physics, Class 12 Video Lecture
| 1. What is ray optics in physics? |  |
Ans. Ray optics is a branch of physics that deals with the behavior of light as it travels through transparent media, such as lenses and mirrors. It focuses on studying the path of light rays and how they interact with different surfaces and materials.
| 2. How does a lens work in ray optics? | |
Ans. A lens works in ray optics by refracting or bending light rays as they pass through it. There are two types of lenses - converging lenses (convex) and diverging lenses (concave). Converging lenses converge the light rays to a focal point, whereas diverging lenses cause the rays to diverge.
| 3. What are the applications of ray optics in real life? | |
Ans. Ray optics has various applications in daily life. Some examples include:
- Glasses and contact lenses: These corrective lenses use the principles of ray optics to correct vision problems.
- Cameras and telescopes: These optical devices use lenses and mirrors to focus light and capture images.
- Microscopes: Microscopes use lenses to magnify small objects and allow us to see them in detail.
- Fibre optics: Fibre optic cables use the principle of total internal reflection to transmit light signals over long distances.
| 4. How does the focal length of a lens affect the image formed? | |
Ans. The focal length of a lens determines the distance at which the image is formed and its characteristics. A lens with a shorter focal length will have a greater bending power and will form images closer to the lens. On the other hand, a lens with a longer focal length will have less bending power and will form images farther away from the lens.
| 5. What is the difference between a real image and a virtual image formed by a lens? | |
Ans. A real image is formed when light rays actually converge at a specific point. It can be projected onto a screen and is always inverted. On the other hand, a virtual image is formed when light rays appear to converge from a specific point, but they do not actually meet. It cannot be projected onto a screen and can be either upright or inverted, depending on the lens.
Text Transcript from Video
hello friends this video on ray optics
part 36 is brought to you by exam feel
calm no more fear from exam please make
sure that you have watched all the
videos to in part 35 before going ahead
with part 36 let us go ahead and look
look at then the problem states that the
image of a small electric bulb fixed on
the wall of a room is to be obtained on
the opposite wall three meters away by
means of a large convex lens so it talks
about a scenario like this let us
suppose this is the wall of a room where
you have a small electric bulb now you
have to obtain the image of this bulb on
the opposite wall which is at a distance
of 3 meters from this wall so your
object lies here and you want the image
to form here by means of a large convex
lens so let us say this is the convex
lens what is the maximum possible focal
length of the lens required for this
purpose so let us say that the maximum
focal length required DF now before we
solve this problem let us talk about an
important concept which we have not have
talked about so far now in this case we
are saying that let us suppose the
object lies here and the image lines
here if this is the object and this is
the image the distance between the image
and the object is given let me call the
distance between the image and the
object as capital D a small D let me
take it as D so D is the distance
between object and image right now if I
apply the thin lens formula applying
thin lens formula we know that
by V minus 1 by U is equal to 1 by F let
us say that the object distances X then
what would be the image distance it
would be D minus X right so this V will
be equal to 1 by D minus X minus 1 by
minus X is equal to 1 by F or we can say
that F is equal to X into D minus X
divided by D so this will be the value
of the focal length now what do I want I
want to find out the maximum value of
focal length for which this D is
obtained so I want the maximum value of
focal length with respect to this
distance right so for that how do we
know the maximum value of focal length
for that we will differentiate this
focal length with respect to this
distance X so DF by DX will should be
equal to 0 this is how we calculate
minimum and maximum values you would
have studied it in your mathematics all
right so the calculate DF by DX so if
you calculate the f by DX DF by DX comes
out to be X minus X square by D is equal
to 0 so D by DX of this because F is
nothing but X minus X square by D
therefore this comes out to be 1 minus 2
X by D is equal to 0 therefore we find
that X is equal to D by 2 there that
means when the value of X is equal to D
by 2 at that time this focal length F is
maximum so f will be maximum when X is
equal to D by 2 therefore F is equal to
let us put the value of x as d by 2
so D by 2 minus D by 2 whole square
divided by D so this comes out to be D
by 4 therefore the maximum value of
focal length is equal to D by 4 where D
is the distance between the object and
the image so now using this formula we
can easily solve this problem because
the distance between the image and
object that is D is given as 3 meters
therefore the maximum focal length will
be equal to D by 4 that is equal to 3 by
4 which is equal to 0.75 meters right so
I hope you understood the concept here
we want to find out the maximum value of
focal lengths what is the value of x for
which the focal length will be maximum
so that is what we calculated with the
help of differentiation so now let us
look at the next problem the problem
says find the position of the image
formed by this lens combination so here
we have a convex lens followed by a
Plano concave lens which is again
followed by a convex lens so let us
consider each lens one by one so what
would be the image formed by first lens
so let us suppose that the image formed
by first lens is i1 so 1 by V 1 minus 1
by u1 is equal to 1 by f1 applying the
thin lens formula so in this case u 1 is
equal to minus 30 centimeters because
the object is located here which is at a
distance of 30 centimeter from the first
lens focal length of the first lens is
given as plus 10 centimeters so putting
these values we can find out 1 by V 1 by
minus 1 by minus 30 is equal to 1 by 10
so we get 1 by V 1 is equal to 1 by 10
minus 1 by 30
or v1 is equal to 15 centimeters right
so the image i1 is formed at a distance
of 15 centimeters from this length right
so that means where is i1 i1 the
situated somewhere here so i1 is located
somewhere somewhere here right now let
us look at so this I 1 act as object for
second lengths right so for the second
lens what would be the object distance u
2 this is the second lens so the object
distance for the second lens would be
this distance z i1 would be located
somewhere here the right eye 1 would be
located somewhere yeah because it is
located at 15 centimeters from the first
lens right so now what is the distance
of I 1 from the second lens that means
this distance so this distance is given
as 10 this distance is given as 10
centimeters therefore u 2 is equal to 10
centimeter and since object is on the
right hand side so this would be
positive and what about the focal length
of the second lens that is given as
minus 10 so again applying the same thin
lens formula 1 by V 2 minus 1 by u 2 is
equal to 1 by F 2 so this implies 1 by V
2 minus 1 by 10 is equal to 1 by minus
10 so this gives the value of V 1 by V 2
as 0 or V 2 comes out to be infinity
that means the image is formed image
formed by the second lens is at infinity
so I 2 is formed at infinity this I 2
acts as object for the third lens that
means for the third lens what is U 3
three is infinity that means the object
is at infinity and what is the focal
length that is given as 30 centimeter so
it is plus 30 centimeters so again the
same thin lens formula 1 by V 3 minus 1
by u 3 is equal to 1 by F 3 so this
implies 1 by V 3 minus 1 by infinity is
equal to 1 by 30 this implies V 3 is
equal to 30 centimeters therefore the
final image so what is the position of
the image formed final image is formed
at 30 centimeter from the third lens so
this would be the position of the final
image does formed right so let us look
at the last problem of this section it
says that if F is equal to 0.5 meter for
a glass lens what is the power of the
lens as discussed we saw that power of a
lens is defined by 1 by focal length so
in this case focal length is given as
0.5 meters so it is 10 by 5 so the focal
length is 2 so the power of the lens is
2 diopters now look at the second part
it says the radii of curvature of the
faces of a double convex lens at 10
centimeters and 15 centimeters that
means R 1 is equal to 10 centimeter R 2
is equal to 15 centimeters its focal
length is 12 centimeters what is the
refractive index of glass Magnus we have
to calculate mu so let us apply the lens
maker formula the lens maker formula
gives 1 by F is equal to MU minus 1 into
1 by r1 minus 1 by r2 therefore why
by f is equal to MU minus 1 into 1 by r1
minus 1 by r2 now in this case R 1 is
positive so r2 has to be negative
because it is a convex lenss so if this
is r1 this is r2
so if r1 is positive this r2 has to be
negative therefore we get mu minus 1
into 1 by 10 plus 1 by 15 is equal to 1
by 12 so from this we can calculate the
value of mu and this value comes out to
be 3 by 2 which is equal to 1.5 thank
you please visit example.com to watch
free educational videos try 3 online
tests get the best quality study
materials study from the best tutors and
mentors and much more thank you once
again
part 36 is brought to you by exam feel
calm no more fear from exam please make
sure that you have watched all the
videos to in part 35 before going ahead
with part 36 let us go ahead and look
look at then the problem states that the
image of a small electric bulb fixed on
the wall of a room is to be obtained on
the opposite wall three meters away by
means of a large convex lens so it talks
about a scenario like this let us
suppose this is the wall of a room where
you have a small electric bulb now you
have to obtain the image of this bulb on
the opposite wall which is at a distance
of 3 meters from this wall so your
object lies here and you want the image
to form here by means of a large convex
lens so let us say this is the convex
lens what is the maximum possible focal
length of the lens required for this
purpose so let us say that the maximum
focal length required DF now before we
solve this problem let us talk about an
important concept which we have not have
talked about so far now in this case we
are saying that let us suppose the
object lies here and the image lines
here if this is the object and this is
the image the distance between the image
and the object is given let me call the
distance between the image and the
object as capital D a small D let me
take it as D so D is the distance
between object and image right now if I
apply the thin lens formula applying
thin lens formula we know that
by V minus 1 by U is equal to 1 by F let
us say that the object distances X then
what would be the image distance it
would be D minus X right so this V will
be equal to 1 by D minus X minus 1 by
minus X is equal to 1 by F or we can say
that F is equal to X into D minus X
divided by D so this will be the value
of the focal length now what do I want I
want to find out the maximum value of
focal length for which this D is
obtained so I want the maximum value of
focal length with respect to this
distance right so for that how do we
know the maximum value of focal length
for that we will differentiate this
focal length with respect to this
distance X so DF by DX will should be
equal to 0 this is how we calculate
minimum and maximum values you would
have studied it in your mathematics all
right so the calculate DF by DX so if
you calculate the f by DX DF by DX comes
out to be X minus X square by D is equal
to 0 so D by DX of this because F is
nothing but X minus X square by D
therefore this comes out to be 1 minus 2
X by D is equal to 0 therefore we find
that X is equal to D by 2 there that
means when the value of X is equal to D
by 2 at that time this focal length F is
maximum so f will be maximum when X is
equal to D by 2 therefore F is equal to
let us put the value of x as d by 2
so D by 2 minus D by 2 whole square
divided by D so this comes out to be D
by 4 therefore the maximum value of
focal length is equal to D by 4 where D
is the distance between the object and
the image so now using this formula we
can easily solve this problem because
the distance between the image and
object that is D is given as 3 meters
therefore the maximum focal length will
be equal to D by 4 that is equal to 3 by
4 which is equal to 0.75 meters right so
I hope you understood the concept here
we want to find out the maximum value of
focal lengths what is the value of x for
which the focal length will be maximum
so that is what we calculated with the
help of differentiation so now let us
look at the next problem the problem
says find the position of the image
formed by this lens combination so here
we have a convex lens followed by a
Plano concave lens which is again
followed by a convex lens so let us
consider each lens one by one so what
would be the image formed by first lens
so let us suppose that the image formed
by first lens is i1 so 1 by V 1 minus 1
by u1 is equal to 1 by f1 applying the
thin lens formula so in this case u 1 is
equal to minus 30 centimeters because
the object is located here which is at a
distance of 30 centimeter from the first
lens focal length of the first lens is
given as plus 10 centimeters so putting
these values we can find out 1 by V 1 by
minus 1 by minus 30 is equal to 1 by 10
so we get 1 by V 1 is equal to 1 by 10
minus 1 by 30
or v1 is equal to 15 centimeters right
so the image i1 is formed at a distance
of 15 centimeters from this length right
so that means where is i1 i1 the
situated somewhere here so i1 is located
somewhere somewhere here right now let
us look at so this I 1 act as object for
second lengths right so for the second
lens what would be the object distance u
2 this is the second lens so the object
distance for the second lens would be
this distance z i1 would be located
somewhere here the right eye 1 would be
located somewhere yeah because it is
located at 15 centimeters from the first
lens right so now what is the distance
of I 1 from the second lens that means
this distance so this distance is given
as 10 this distance is given as 10
centimeters therefore u 2 is equal to 10
centimeter and since object is on the
right hand side so this would be
positive and what about the focal length
of the second lens that is given as
minus 10 so again applying the same thin
lens formula 1 by V 2 minus 1 by u 2 is
equal to 1 by F 2 so this implies 1 by V
2 minus 1 by 10 is equal to 1 by minus
10 so this gives the value of V 1 by V 2
as 0 or V 2 comes out to be infinity
that means the image is formed image
formed by the second lens is at infinity
so I 2 is formed at infinity this I 2
acts as object for the third lens that
means for the third lens what is U 3
three is infinity that means the object
is at infinity and what is the focal
length that is given as 30 centimeter so
it is plus 30 centimeters so again the
same thin lens formula 1 by V 3 minus 1
by u 3 is equal to 1 by F 3 so this
implies 1 by V 3 minus 1 by infinity is
equal to 1 by 30 this implies V 3 is
equal to 30 centimeters therefore the
final image so what is the position of
the image formed final image is formed
at 30 centimeter from the third lens so
this would be the position of the final
image does formed right so let us look
at the last problem of this section it
says that if F is equal to 0.5 meter for
a glass lens what is the power of the
lens as discussed we saw that power of a
lens is defined by 1 by focal length so
in this case focal length is given as
0.5 meters so it is 10 by 5 so the focal
length is 2 so the power of the lens is
2 diopters now look at the second part
it says the radii of curvature of the
faces of a double convex lens at 10
centimeters and 15 centimeters that
means R 1 is equal to 10 centimeter R 2
is equal to 15 centimeters its focal
length is 12 centimeters what is the
refractive index of glass Magnus we have
to calculate mu so let us apply the lens
maker formula the lens maker formula
gives 1 by F is equal to MU minus 1 into
1 by r1 minus 1 by r2 therefore why
by f is equal to MU minus 1 into 1 by r1
minus 1 by r2 now in this case R 1 is
positive so r2 has to be negative
because it is a convex lenss so if this
is r1 this is r2
so if r1 is positive this r2 has to be
negative therefore we get mu minus 1
into 1 by 10 plus 1 by 15 is equal to 1
by 12 so from this we can calculate the
value of mu and this value comes out to
be 3 by 2 which is equal to 1.5 thank
you please visit example.com to watch
free educational videos try 3 online
tests get the best quality study
materials study from the best tutors and
mentors and much more thank you once
again
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Nov 25, 2024 Last updated |
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