Light: Refraction and Reflection || Exercise Question || CBSE/NCERT
Light is a form of
energy which produced in us the sensation of vision. Speed of light in vacuum
is 3 x 10^8 meter per second. White light of sunlight consist of seven colors.
Ques.1:_ Which one of the
following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:_ A
lens allow light to pass through it since clay does not show such properties; it
cannot be used to make a lens.
Ques.2:_ The image formed by a concave
mirror is observed to be virtual, erect and larger than the object. Where
should be the position of the object?
(a)
Between the principal focus and the centre of curvature.
(b)
At the centre of curvature.
(c)
Beyond the centre of curvature.
(d)
Between the pole of the mirror and its principal focus.
Answer:_
It should be between the principal focus and centre of
curvature.
Question 3:_ Where should an object
be placed in the front of convex lens to get a real image of the size of the
object?
(a)
At the principal focus of lens.
(b)At
twice the focal length.
(c)
At Infinity
(d)
Between the optical centre of the lens and its principal axis.
Answer:_ (b) At twice the focal
length i.e. at the centre of curvature.
Question 4:_ A spherical mirror and
a thin spherical lens have each a focal length ─15 cm. The mirror and the lens are
likely to be:-
(a)
both concave
(b)
Both convex
(c)
the mirror is concave and lens is convex.
(d)
the mirror is convex but the lens is concave.
Answer:_
(a) both concave. As focal length of the concave lens and a
concave mirror are taken as negative.
Question 5:_ No matter how far
stand from a mirror, your image appears erect. The mirror is likely to be:
(a)
Plane
(b)
Concave
(c)
Convex
(d)
either plane or convex.
Answer:
_
(d) either plane or convex.
Question 6:_ Which of the following
lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b)
A concave lens of focal length 50 cm.
(c)
A convex lens of focal length 5 cm.
(d)
A concave lens of focal length 5 cm
Answer:_ A
convex lens of focal length 5 cm would prefer to use while reading small letters
found in dictionary.
Question 7:_ We wise to obtain
an erect image of an object, using a concave mirror of focal length 15 cm. What
should be the range of the distance of the object from the mirror? What is the
nature of the image? Is the image larger or smaller than the object? Draw a ray
diagram to show the image formation in this case.
Answer:_ Range of object distance = 0 cm to 15 cm.
A
concave mirror give an erect image when an object is placed between its pole
(P) and the principal focus.
Hence,
to obtain an erect image of an object from a concave mirror of focal length 15 cm,
the object must be placed anywhere between the pole and the focus. The image
formed will be virtual erect and magnified in nature as shown in figure.
Question 8:_ Name the type of
mirror used in the following situation:
(a)
Headlight of a car
(b)
Side/rear-view mirror of a vehicles.
(c)
Solar furnace
Answer:_
(a) Headlight of a car: Concave
Reason:
Concave mirror is used in headlight of a car. This is because concave mirror
can produced powerful parallel a beam of light. When the light source is placed
at their principal focus.
(b)
Side/rear-view mirror of vehicles: Convex
Reason: Convex mirror give a
virtual, erect and diminished image of the object placed in front of it.
Because of this, they have a wide field of view. It enables the driver to see
most of the traffic behind him/her.
(c)
Solar furnace: Concave
Reason:
Concave mirror are convergent mirrors. That is why they are used to construct
solar furnaces. Concave mirrors converge the light incident on them at a single
point known as principal focus. Hence they can be used to produce large amount
of heat at that point.
Question 9:_ One half of a convex
lens is covered with a black paper. Will this lens produce a complete image of
an object? Verify your answer experimentally. Explain your observations.
Answer:_
The convex lens will form complete image of an object, even if
its one half is covered with black paper. It can be understood by the following
two cases.
Case I:
When the upper half of the lens is covered.
In
this case a ray of light coming from the object will be reflected by the lower
half of the lens. These rays meet at the other side of the lens to form the image
of the given object as shown in following figure.
Case II: When the lower half of
lens is convered.
In
this case, a ray of light coming from the object is refracted by the upper half
of the lens. These rays meet at the other side of the lens to form the image of
the given object as shown in the following figure.
Question 10:_ An object 5 cm in
length is held 25 cm away from a converging lens of focal length 10 cm. Draw
the ray diagram and find the position, size and nature of the image formed.
Answer:_
Object distance u = ─ 25 cm
Object
height = 5 cm
Focal
length f = + 10 cm
According
to lens formula
cm
The
position value of v shows that the image is formed at the other side of the lens.
The
negative sign shows that the image is real and formed behind the lens.
H1
= m × Ho = ─ 0•66
×
5 = ─ 3•3 cm.
The
negative value of image height indicate that image formed in inverted. The
position, size and nature of image are
shown in following ray diagram.
Question 11:_ A concave lens of
focal length 15 cm from an image 10 cm from the lens. How far is the object
place from the lens? Draw the diagram.
Answer:_
Focal length of concave lens ( OF1) f = ─ 15 cm
Image
distance, v = ─ 10 cm
According
to lens formula
The
negative value of the u indicates that the object is
placed 30 cm in front of the lens. This is shown in the following ray diagram.
Question 12:_ An object is placed at
a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position
and nature of the image.
Answer:_
Focal length of convex mirror, ( f ) = + 15 cm
Object
distance (u) = ─ 10 cm
According
to mirror formula
The
+ve value of v indicates that the image formed behind the mirror.
m = + 0・6
The
+ve value of magnification indicates
that image formed is virtual and erect.
Answer:_
We know that
Since
the magnification produced is +1. It means the image formed by the plane mirror
is of the same size as that of the object. The positive side show that image
formed is virtual and erect.
Question 14:_ An object 5•0 cm in length is placed at a distance of 20 cm in
front of a convex mirror of radius of curvature 30 cm. Find the position of the
image, its nature and size.
Answer:_
Object distance (u) = ─ 20 cm
Object
Height (h) = ─ 5 cm
Radius
of curvature (R) = ─30 cm
Since, Radius of curvature = 2 ×
Focal Length
\
R = 2 × f
⟹ f = 30
⁄ 2 = 15 cm i.e f = + 15 cm
According
to mirror formula
\ v = 8・57 cm
The
positive value of v indicates that the mirror is formed behind the mirror.
h/ = m × h = 0•428 × 5 = 2•14 cm ⟹ h/ = 2•14 cm
The positive value of magnification indicate that the image formed is erect. Therefore the image formed is virtual, erect and smaller in size.
Question 15:_ An object of size 7 cm
is placed at 27 cm in front of the concave mirror of focal length 18 cm. At
what distance from the mirror should a screen be placed. So that sharp focused
image can be obtained? Find the size and
nature of the image.
Answer:_
Object distance (u) = ─ 27 cm
Object
Height (h) = 7 cm
Focal
Length f = ─ 18 cm
According
to mirror formula
⟹ v = ─ 54 cm
The screen should be placed at a distance of 54 cm in
front of the given mirror.
The negative value of magnification indicates that the
image formed is real.
h/ = m × h = 7
× ─ 2 = ─ 14 cm ⟹
h/ = ─
14 cm
The negative sign show that image formed is inverted.
Question
16:_ Find the
focal length of power ─ 20 D. What types of lens is this?
Answer:_
A concave lens has a negative focal length. Hence it is a concave lens.
Question
17:_ A doctor
has prescribed a corrective lens of power + 1•5 D. Find the focal length of the lens. Is the
prescribed lens diverging or converging.
Answer:_
Hence
the lens is convex or a converging lens.
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