CBSE Class 10 > Class 10 Videos > Numericals - Light Reflection & Refraction, CBSE Class 10 Physics
Numericals - Light Reflection & Refraction, CBSE Class 10 Physics Video
FAQs on Numericals - Light Reflection & Refraction, CBSE Class 10 Physics
| 1. What is light reflection and refraction? |  |
Ans. Light reflection refers to the bouncing back of light when it strikes a surface. Refraction, on the other hand, is the bending of light as it passes from one medium to another. These phenomena play a crucial role in understanding how light behaves and interacts with different materials.
| 2. How does light reflection occur? | |
Ans. Light reflection occurs when light rays strike a surface and rebound in various directions. The angle at which the light ray strikes the surface is called the angle of incidence, and the angle at which it bounces off is known as the angle of reflection. The law of reflection states that the angle of incidence is always equal to the angle of reflection.
| 3. What is the difference between regular and diffuse reflection? | |
Ans. Regular reflection occurs when light rays strike a smooth and polished surface, such as a mirror. In this case, the light rays bounce off at the same angle as the angle of incidence, resulting in a clear and sharp image. Diffuse reflection, on the other hand, happens when light rays strike an irregular or rough surface. The light is scattered in different directions, causing a blurred or scattered reflection.
| 4. How does light refraction occur? | |
Ans. Light refraction occurs when light passes from one medium to another, such as from air to water or from air to glass. As light travels through different mediums, its speed and direction change, causing it to bend. This bending of light is due to the change in its velocity caused by the change in the density of the medium.
| 5. What are some everyday examples of light reflection and refraction? | |
Ans. Light reflection and refraction are present in our everyday lives. Some examples of reflection include seeing your reflection in a mirror, the shiny surface of a car, or the reflection of light from a still water surface. Examples of refraction include the bending of a pencil in a glass of water, the formation of a rainbow when sunlight passes through raindrops, and the apparent bending of a straw in a glass of water.
Text Transcript from Video
hello friends this video on light
reflection and refraction part 12 is
brought to you by exam comm no more fear
from exam please make sure that you have
watched all the videos - part 11 before
going ahead with part 12 turn all this
let us try to solve some problems the
problems is that a convex mirror has its
radius of curvature is equal to 20
centimeters find the position of the
image of an object placed at a distance
of 12 centimeter from the mirror so let
us run this is my convex mirror convex
means the outside of the spherical
surfaces silvered right so you have an
object placed in front of this mirror
the radius of curvature that is capital
R is given as 20 centimeters so what
would be the radius of curvature of this
mirror it would be somewhere here right
so it is on the right hand side so it
will be positive find the position of
the image of an object placed at a
distance of 12 centimeter from that
mirror now object is always placed on
the left hand side of the middle right
and all distances will be measured from
pole so that means this is my object
distance view and the object distance is
12 centimeters since it is on the left
hand side of the pole so it would be
negative so we have to find out the
position of the image that means we will
have to calculate the value of V so we
will make use of mirror equations for
mirror equation we want you V and F now
we know a relationship between focal
length and radius of curvature what is
that relationship radius of curvature is
twice the focal length so we can say
that focal length will be equal to R by
2 that is 20 by 2 which is equal to plus
10 centimeters now we will make use of
middle equation so as per the mirror
equation 1 plus 1 by u plus 1 by V is
equal to 1 by F
so now let us put the values of U and F
in this equation so that we can
calculate the value of U V so we get 1
by 10 minus 1 by minus 12 so this is
equal to 1 by 10 plus 1 by 12 so this
comes out to be 11 by 60 or we can say V
is equal to 60 by 11 centimeters so the
value of V comes out to be positive that
means the image is formed somewhere here
that means somewhere on the right-hand
side of the pole
okay so let us look at the next problem
it says that a small candle 2 point 5
centimeter in size is placed at 27
centimeters in front of a concave mirror
of radius of curvature 36 centimeters
that means you have a concave mirror
somewhat like this you have a small
candle the height of the candle is given
as 2.5 centimeters that means height of
the object is 2.5 centimeters now since
it is above the principal axis so this
would be positive and it is placed at 27
centimeters in front of the concave
mirror that means the object distance is
27 centimeters it is on the left hand
side of the pole therefore it is
negative the radius of curvature is
given as 36 centimeters so the radius of
curvature for the sphere will also fall
on this side so this would be negative
right so we have to calculate at what
distance from the mirror should a stream
be placed in order to obtain a sharp
image so where should you place the
screen so that you get a sharp image you
should calculate the distance at which
the image will be formed so that you can
place the screen at that location itself
that means you have to guide you take
the value of V and you have to also tell
about the nature and
of the image so let us first try to
calculate V using the middle equation
now using mirror equation we know that 1
by u plus 1 by V is equal to 1 by F now
how do we calculate F F is nothing but R
divided by 2 so from that we can say
that F is nothing but minus 18
centimeters so we will put that value in
this expression so we get 1 by V is
equal to 1 by F minus 1 by u so that is
if working 1 by minus 18 minus 1 by
minus 27 so from this we get V is equal
to minus 54 centimeters so what does it
say it says that the image will be
formed at a distance of 54 centimeters
from the mirror and the sign is negative
that means the image will be formed
somewhere here right this would be my
image now it doesn't tell anything about
whether the image will be erect virtual
or how will it be so for nature and size
of the image in order to know the nature
of the image and size of the image let
us try to calculate the magnification in
this case what is magnification
magnification can be calculated as the
ratio of image distance by object
distance right so in this case what is
my image distance it is minus 54 what is
my object distance it is minus 27 so the
magnification comes out to be minus 2
also we know that magnification can be
defined as the height of the image by
height of the object now
magnification is minus 2 height of image
we do not know and height of the object
is given as 2.5 centimeters so from this
we can say H dash is equal to minus 5
centimeters so from this what can we
tell about the nature of the image since
H dash is negative so we can say that
the email
is inverted since it is five centimeters
so it is magnified or it is enlarged
now since the image is inverted so it
has to be a real image that means how
good be the image this will not be the
image the image will be somewhat like
this it will be enlarged it would be
inverted and real right so we calculated
the nature size and position of the
image let us now look at the third
problem it says a concave middle
produces three times magnified or
enlarged real image of an object placed
at ten centimeters in front of it where
is the image located that means in this
thing's
the magnification is given as three
because it produces three times enlarged
image that means height of the image by
height of the object is equal to three
that is how we define magnification
height of image by height of object that
is given as three we have to calculate
the image distance the object distance
is given as ten centimeters now object
is always placed on left-hand side of
the pole so it would be negative right
so now from definition of magnification
we can say that magnification is equal
to the image distance divided by the
object distance now here it says that it
produces a real image that means it the
real image is always inverted so your
magnification should be negative because
H dash will be negative right so this
magnification becomes minus three so now
we know that magnification is equal to
minus V by u so we can write
magnification as minus three this
becomes minus V divided by U is minus
ten so from this we can say that magnet
V comes out to remind is thirty
centimeters that means the image will be
formed at a distance of thirty
centimeters from the pole and it will be
formed towards the left hand side of the
pole right so please take care that put
sign convention
consistently here it talked about real
image so you should put the
magnification as negative because it
talked about real image all these means
and inverted image virtual image would
mean a direct image so in that case we
will put magnification is positive let
us look at the fourth problem it says an
object of length two point five
centimeters is placed at a distance of
1.5 F from a concave mirror where F is
the magnitude of the focal length of the
mirror find the length and nature of the
image so in this case let us assume that
you have a mirror like this upon cave
mirror length of the object that is
height of the object is given as two
point five centimeters distance of the
image U is given as 1 point five F now
since the object is placed on left-hand
side so it would be negative focal
length of the mirror is f find the
length and nature of the image so we
have to calculate H - and also we have
to calculate the nature of the image so
for that we will also have to calculate
V so let us make use of mirror equation
so from mirror equation 1 by u plus 1 by
G is equal to 1 by F so what is U it is
minus 1.5 F plus 1 by V is equal to 1 by
F now what would be the value of focal
method focus focus will also be on the
left hand side so this will also be
minus F so from this we can say 1 by V
is equal to 1 by minus F plus 1 by 1.5 F
so this becomes minus 0.5 by 1.5 F so
from this we can say
is equal to minus three F so this is the
binding for image distance right now
using this image distance we can
calculate magnification which is defined
as minus V by u so magnification will be
equal to minus minus three F divided by
u is minus 1.5 F so this minus minus
will get cancelled so this will get
cancelled we get minus 2 F and F will
also get cancelled so magnification is
minus 2 so once I know the magnification
and the height of the object I can
calculate the height of the image
magnification is defined as height of
image by height of object so I can
calculate height of image as
magnification into height of object so
what was the magnification calculated it
is minus 2 and what is the height of the
object it is 2.5 centimeters so this
will be minus 2 into 2.5 centimeters
which is minus 5 centimeters so we
calculated the size of the object find
the length of the object length we have
calculated and what would be the nature
of the image nature would be inverted so
if it is inverted that means it is real
and the size is enlarged so the nature
of the image is real inverted and
enlarged and the size of the images 5
centimeters let us look at the last
problem of this topic it says we wish to
obtain an erect image of an object using
upon cave mirror of focal length 15
centimeters okay what should be the
range of distance of the object from the
middle I mean for concave mirror you
would have seen we plot it for different
object distances right so we can get a
real and inverted images we can also get
virtual and erect images but what did we
see we saw that when the object was play
very close to the concave mirror that
means when the object was placed between
the pole and principal focus if this is
the pole and this is the principal focus
only when the object was placed between
pole and principal focus we got a
virtual and erect image because in that
case something like this happened and
therefore these sorry therefore these
two rays met at this point and so we got
a virtual and erect image so for a
virtual image the object should be
placed object should be placed between
principal focus and the pole that is
them that should be the range of
distance right is the image larger or
smaller than the object the image is
larger than the object image is larger
than object draw a ray diagram to show
the image formation in this case so that
I have already drawn me right this line
was drawn by mistake please try to plot
all the ray diagrams yourselves so that
you can get an idea that how and where
exactly the image is getting formed
right thank you please visit exam fee or
comm to watch free educational videos
try free online tests get the best
quality study materials study from the
best tutors and mentors and much more
thank you once again
reflection and refraction part 12 is
brought to you by exam comm no more fear
from exam please make sure that you have
watched all the videos - part 11 before
going ahead with part 12 turn all this
let us try to solve some problems the
problems is that a convex mirror has its
radius of curvature is equal to 20
centimeters find the position of the
image of an object placed at a distance
of 12 centimeter from the mirror so let
us run this is my convex mirror convex
means the outside of the spherical
surfaces silvered right so you have an
object placed in front of this mirror
the radius of curvature that is capital
R is given as 20 centimeters so what
would be the radius of curvature of this
mirror it would be somewhere here right
so it is on the right hand side so it
will be positive find the position of
the image of an object placed at a
distance of 12 centimeter from that
mirror now object is always placed on
the left hand side of the middle right
and all distances will be measured from
pole so that means this is my object
distance view and the object distance is
12 centimeters since it is on the left
hand side of the pole so it would be
negative so we have to find out the
position of the image that means we will
have to calculate the value of V so we
will make use of mirror equations for
mirror equation we want you V and F now
we know a relationship between focal
length and radius of curvature what is
that relationship radius of curvature is
twice the focal length so we can say
that focal length will be equal to R by
2 that is 20 by 2 which is equal to plus
10 centimeters now we will make use of
middle equation so as per the mirror
equation 1 plus 1 by u plus 1 by V is
equal to 1 by F
so now let us put the values of U and F
in this equation so that we can
calculate the value of U V so we get 1
by 10 minus 1 by minus 12 so this is
equal to 1 by 10 plus 1 by 12 so this
comes out to be 11 by 60 or we can say V
is equal to 60 by 11 centimeters so the
value of V comes out to be positive that
means the image is formed somewhere here
that means somewhere on the right-hand
side of the pole
okay so let us look at the next problem
it says that a small candle 2 point 5
centimeter in size is placed at 27
centimeters in front of a concave mirror
of radius of curvature 36 centimeters
that means you have a concave mirror
somewhat like this you have a small
candle the height of the candle is given
as 2.5 centimeters that means height of
the object is 2.5 centimeters now since
it is above the principal axis so this
would be positive and it is placed at 27
centimeters in front of the concave
mirror that means the object distance is
27 centimeters it is on the left hand
side of the pole therefore it is
negative the radius of curvature is
given as 36 centimeters so the radius of
curvature for the sphere will also fall
on this side so this would be negative
right so we have to calculate at what
distance from the mirror should a stream
be placed in order to obtain a sharp
image so where should you place the
screen so that you get a sharp image you
should calculate the distance at which
the image will be formed so that you can
place the screen at that location itself
that means you have to guide you take
the value of V and you have to also tell
about the nature and
of the image so let us first try to
calculate V using the middle equation
now using mirror equation we know that 1
by u plus 1 by V is equal to 1 by F now
how do we calculate F F is nothing but R
divided by 2 so from that we can say
that F is nothing but minus 18
centimeters so we will put that value in
this expression so we get 1 by V is
equal to 1 by F minus 1 by u so that is
if working 1 by minus 18 minus 1 by
minus 27 so from this we get V is equal
to minus 54 centimeters so what does it
say it says that the image will be
formed at a distance of 54 centimeters
from the mirror and the sign is negative
that means the image will be formed
somewhere here right this would be my
image now it doesn't tell anything about
whether the image will be erect virtual
or how will it be so for nature and size
of the image in order to know the nature
of the image and size of the image let
us try to calculate the magnification in
this case what is magnification
magnification can be calculated as the
ratio of image distance by object
distance right so in this case what is
my image distance it is minus 54 what is
my object distance it is minus 27 so the
magnification comes out to be minus 2
also we know that magnification can be
defined as the height of the image by
height of the object now
magnification is minus 2 height of image
we do not know and height of the object
is given as 2.5 centimeters so from this
we can say H dash is equal to minus 5
centimeters so from this what can we
tell about the nature of the image since
H dash is negative so we can say that
the email
is inverted since it is five centimeters
so it is magnified or it is enlarged
now since the image is inverted so it
has to be a real image that means how
good be the image this will not be the
image the image will be somewhat like
this it will be enlarged it would be
inverted and real right so we calculated
the nature size and position of the
image let us now look at the third
problem it says a concave middle
produces three times magnified or
enlarged real image of an object placed
at ten centimeters in front of it where
is the image located that means in this
thing's
the magnification is given as three
because it produces three times enlarged
image that means height of the image by
height of the object is equal to three
that is how we define magnification
height of image by height of object that
is given as three we have to calculate
the image distance the object distance
is given as ten centimeters now object
is always placed on left-hand side of
the pole so it would be negative right
so now from definition of magnification
we can say that magnification is equal
to the image distance divided by the
object distance now here it says that it
produces a real image that means it the
real image is always inverted so your
magnification should be negative because
H dash will be negative right so this
magnification becomes minus three so now
we know that magnification is equal to
minus V by u so we can write
magnification as minus three this
becomes minus V divided by U is minus
ten so from this we can say that magnet
V comes out to remind is thirty
centimeters that means the image will be
formed at a distance of thirty
centimeters from the pole and it will be
formed towards the left hand side of the
pole right so please take care that put
sign convention
consistently here it talked about real
image so you should put the
magnification as negative because it
talked about real image all these means
and inverted image virtual image would
mean a direct image so in that case we
will put magnification is positive let
us look at the fourth problem it says an
object of length two point five
centimeters is placed at a distance of
1.5 F from a concave mirror where F is
the magnitude of the focal length of the
mirror find the length and nature of the
image so in this case let us assume that
you have a mirror like this upon cave
mirror length of the object that is
height of the object is given as two
point five centimeters distance of the
image U is given as 1 point five F now
since the object is placed on left-hand
side so it would be negative focal
length of the mirror is f find the
length and nature of the image so we
have to calculate H - and also we have
to calculate the nature of the image so
for that we will also have to calculate
V so let us make use of mirror equation
so from mirror equation 1 by u plus 1 by
G is equal to 1 by F so what is U it is
minus 1.5 F plus 1 by V is equal to 1 by
F now what would be the value of focal
method focus focus will also be on the
left hand side so this will also be
minus F so from this we can say 1 by V
is equal to 1 by minus F plus 1 by 1.5 F
so this becomes minus 0.5 by 1.5 F so
from this we can say
is equal to minus three F so this is the
binding for image distance right now
using this image distance we can
calculate magnification which is defined
as minus V by u so magnification will be
equal to minus minus three F divided by
u is minus 1.5 F so this minus minus
will get cancelled so this will get
cancelled we get minus 2 F and F will
also get cancelled so magnification is
minus 2 so once I know the magnification
and the height of the object I can
calculate the height of the image
magnification is defined as height of
image by height of object so I can
calculate height of image as
magnification into height of object so
what was the magnification calculated it
is minus 2 and what is the height of the
object it is 2.5 centimeters so this
will be minus 2 into 2.5 centimeters
which is minus 5 centimeters so we
calculated the size of the object find
the length of the object length we have
calculated and what would be the nature
of the image nature would be inverted so
if it is inverted that means it is real
and the size is enlarged so the nature
of the image is real inverted and
enlarged and the size of the images 5
centimeters let us look at the last
problem of this topic it says we wish to
obtain an erect image of an object using
upon cave mirror of focal length 15
centimeters okay what should be the
range of distance of the object from the
middle I mean for concave mirror you
would have seen we plot it for different
object distances right so we can get a
real and inverted images we can also get
virtual and erect images but what did we
see we saw that when the object was play
very close to the concave mirror that
means when the object was placed between
the pole and principal focus if this is
the pole and this is the principal focus
only when the object was placed between
pole and principal focus we got a
virtual and erect image because in that
case something like this happened and
therefore these sorry therefore these
two rays met at this point and so we got
a virtual and erect image so for a
virtual image the object should be
placed object should be placed between
principal focus and the pole that is
them that should be the range of
distance right is the image larger or
smaller than the object the image is
larger than the object image is larger
than object draw a ray diagram to show
the image formation in this case so that
I have already drawn me right this line
was drawn by mistake please try to plot
all the ray diagrams yourselves so that
you can get an idea that how and where
exactly the image is getting formed
right thank you please visit exam fee or
comm to watch free educational videos
try free 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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Apr 18, 2026 Last updated
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