REFLECTION OF LIGHT
Exercise 7A
Numerical
1.A ray of light is incident on a plane mirror.Its reflected ray
is perpendicular to the incident ray.Find
the angle of incidence?
Answer:
Angle of incidence (i) + Angle of reflection(r) = 90o
But, according to the laws of reflection, i
= r
Therefore, 2 i = 90o
i = r = 45o
2. A man standing in front of a plane mirror finds his image at
a distance 6 meter from himself. What
is the distance of man from the mirror?
Answer:
Distance between man and his image = 6 m
Distance between man and mirror + distance
between mirror and image = 6m
But, Distance between man and mirror =
distance between mirror and image
Therefore, distance of man from mirror =
6/2 = 3 m
3. An insect is sitting
in front of a plane mirror at a distance 1m from it.
(a) Where is the image of the insect formed?
(b) What
is the distance between the insect and its image?
Answer:
(a) Image of the insect is formed 1m behind
the mirror.
(b) Distance between the insect and his image
= 1 + 1 = 2 m
4. An object is kept at 60 cm
in front of a plane mirror.If the mirror is now moved 25 cm
away from
the object,how does the image shift from its previous position?
Answer:
Distance of the object from the mirror = 60
cm.
Therefore, image is formed at a distance 60
cm from the mirror, behind it.
Thus, initial distance between the object
and image = 60 + 60 = 120 cm
If the mirror is moved 25 cm away from the
object,the new distance of the
object from the mirror = 60 + 25 = 85 cm
The
new image is now at a distance 85 cm from the mirror behind it.
Thus,
new distance of the image from the object = 85 + 85 = 170 cm
Taking
the position of the object as reference point, the distance between the
two positions of the image = new distance
of image from the object - initial
distance of the image from the object = (170 - 120) cm = 50 cm
Thus,
the image shifts 50 cm away.
5. An optician while testing
the eyes of a patient keeps a chart of letters 3 m behind the
patient and
asks him to see the letters on the image of chart formed in a plane mirror kept at distance 2 m
in front of him.At
what distance is the chart seen by the
patient?
Answer:
Distance between man and chart = 3 m
Distance between man and mirror = 2 m
Therefore, distance between chart and
mirror = 5 m
Now, final image is formed on the mirror,
which is at a distance of 2 m from the man,
therefore,
the chart as seen by patient is (5 m + 2 m =) 7 m away.
the chart as seen by patient is (5 m + 2 m =) 7 m away.
Exercise 7 (B)
Numerical
Numerical
Numerical
1. State the number of images of an object placed between the
two plane mirrors,formed in each
case
when the mirrors are inclined to each other at
(a) 90o and (b) 60o
Answer:
(a) Angle between the mirrors, ፀ = 90o
Now, n = 360o / ፀ = 360o / 90o = 4, which
is even.
Hence number of images formed
will be (n-1); i.e., 4-1 = 3 images
(b) Angle between the mirrors, ፀ = 60o
Now, n = 360o / ፀ = 360o / 60o = 6, which
is even.
2. An Object is placed (i) asymmetrically (ii)
symmetrically, between two plane mirrors inclined
at an angle of 50o.Find the number of images formed?
Exercise 7(C)at an angle of 50o.Find the number of images formed?
Answer:
Angle between
the mirrors, = 50o
= 50o
Now, n = 360o / ፀ = 360o / 50o = 7.2 
7, which is odd.
7, which is odd.
(i) When
placed asymmetrically, number of images formed will be n, i.e. 7.
(ii) When
placed symmetrically, number of images formed will be (n-1);
i.e. 7-1 = 6
images
Numerical
1. The radius of curvature of a convex mirror is 40 cm.Find its
focal length?
Answer:
Focal length = ½ (Radius of curvature) Or, f = 40/2 = 20 cm
2. The focal length of a concave mirror is 10 cm. Find its
radius of curvature.
Answer:
Radius of curvature = 2 focal length
focal length
Or, R = 2f
= 2 10 = 20 cm
10 = 20 cm
3. An object of height 2 cm is placed at a distance 20 cm in
front of a concave mirror of focal length
12 cm. Find the position,size and nature of the image.
Answer:
The
image is 30 cm in front of the mirror, 3 cm high, real, inverted and magnified.
4. An object is placed at 4 cm distance in
front of a concave mirror of radius of curvature
24 cm.Find
the position of image.Is the image magnified?
Answer:
The image is 6
cm behind the mirror.
Yes. the image is magnified.
5. At what distance from a concave of focal length 25 cm should an
object be
placed so that the size of image is equal to the size of the
object?
Answer:
The size of the image is equal to the size of
the object if the object is placed at the
centre of
curvature of a concave mirror.Hence, the object should be placed at 50 cm.
6.An object 5 cm high is placed at a distance 60 cm in front of a concave mirror of focal length 10
cm.Find (i) the position and (ii) size of the image.
cm.Find (i) the position and (ii) size of the image.
The position of the object is 12 cm in front of the mirror.
Its size is 1 cm.
Its size is 1 cm.
7. A point light source is kept in front of a convex mirror at a
distance of 40 cm.The focal length
of the mirror is 40 cm. Find the position of the image.
Answer:
The image is behind the mirror at a distance 20 cm.
8. When an object of height 1 cm is kept at a distance 4 cm from a
concave mirror,its erect image
of height 1.5 cm is
formed at a distance 6 cm behind the mirror.
Find the focal length of
mirror
Answer:
A ray passing
parallel to the principal axis passes through the focal point after reflection. Hence, the
focal length is 12 cm.
focal length is 12 cm.
9. An object of length 4 cm is placed in front of a concave mirror
at a distance 30 cm.The focal
length of mirror is 15 cm. (a)
Where will the image form?
(b) What will be the
length of the image?
Answer:
O=4 cm ,u=-30
cm ,f=-15 cm
from mirror formula 1 / f= 1 / u + 1 / v
1 / v=1 / f - 1 / u = 1 / -15 -
1 / -30 = 1 / 30 - 1 / 15
1 / v = - 1 / 30 ,v = -30
cm
Hence the image is formed at a
distance of 30 cm in front of the mirror.
m= - v / u= I / O
I = - Ov / u = -4* -30 / -30 = -4 cm
Negative sign indicates inverted image.So I = 4
cm. The length of the image is 4 cm.
10. A concave mirror forms a real image of an object placed in
front of it at a distance
30
cm,of size three times the size of object. Find (a) the focal length of mirror
(b) position
of image
Answer:
u= -30 cm
m= I / O = 30 / O = 3
But , for real object m is negative
m= -3
m= - ( v / u ) , ∴
- (v / u ) = -3
v= 3u
v= 3 x -30 = -90 cm
So the position of image is 90 cm in front
of the mirror.
From
mirror formula,
1/f =
1/v + 1/u
∴ 1/f = 1/
-90 + 1 / -30 = -(1 / 90 )- (1 / 30 )=( - 1 -3 ) / 90 = -4 / 90
f = -22.5 cm
11. A concave mirror forms a virtual image of size twice that of the object placed at a distance 5 cm
from it.Find :(a) the focal length of the mirror (b) position of image.
Answer:
m = I / O = 20 / 2 = 2
m= -(v / u)
∴ -(v / u) = 2
-v = 2 u
u= -5 cm
v = 10 cm
So the position of image is 10 cm in behind the mirror.
from it.Find :(a) the focal length of the mirror (b) position of image.
Answer:
m = I / O = 20 / 2 = 2
m= -(v / u)
∴ -(v / u) = 2
-v = 2 u
u= -5 cm
v = 10 cm
So the position of image is 10 cm in behind the mirror.
From mirror formula,
1/f = 1/v + 1/u
∴ 1/f = 1/ 10 + 1 / -5 =(1 / 10) - (1 / 5)
f= -10 cm
1/f = 1/v + 1/u
∴ 1/f = 1/ 10 + 1 / -5 =(1 / 10) - (1 / 5)
f= -10 cm
12.The image formed by a concave mirror is of size double the size of object. How are u and v
related?
Answer:
magnification is
related?
Answer:
magnification is
m = I / O = (1/3 O) / O = 1 / 3
A convex mirror always forms a virtual and upright image. So ,
m= -v / u
-v / u =1 / 3
v = - 1/ 3 u
But , u is always negative
v = 1 / 3 u
u = -3v
13.The erect image formed by a concave mirror is of size double the size of object.How are u and v
related?
Answer:
magnification is
m = I / O = 2 O / O = 2
m = - (v / u)
-(v / u) = 2
v = -2u
But , u is always negative
v = 2u
14.The magnification for a mirror is -3.How are u and v related?
Answer:
Magnification of a mirror is
m = - v / u
m = -3
-3 = - (v / u )
v = 3 u
But , u will always be negative
v = -3 u