LIGHT 2ND - PHYSICS LIGHT - SOLUTION - e

File No.33/07/18/11/2014
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
ANDHRA PRADESH - TELANGANA
2014-2015 PROGRAM M E
2ND - PHYSICS LIGHT - SOLUTION
LIGHT
1.
The sideways reversal of the image by plane mirror is virtual and erect.
2.
When you look at yourself in a plane mirror, you see a virtual image that appears to be behind the mirror
3.
You are watching your right palm in the plane mirror by placing it parallel to it. then you will observe your
palm as Left palm
4.
Cameras, Microscopes and Eye glasses .
5.
A
B
i
O
Plane mirror
From diagram, and “OP” represents, normal to the plane mirror
6.
Periscope is employed to clearly view objects
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cannot be seen directly due to obstruction.
7.
The nature of the image formed by a plane mirror is virtual and erect, of the same size as the object and
laterally inverted
8.
A ray of light is incidence at an angle of 350 on a plane mirror is
Plane mirror
0

35
Incident ray
Re flected ray
Incident ray = Reflected ray
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 =350
9.
Oiled paper allows only a part of the light to pass through it
10.
Coming back of light into same medium from a boundary seperating two media is called Reflection
11.
Angle of incidence is 00, angle of reflection is 00 & angle of deviation is1800 is a light incident along normal
12.
Given  = 450
angle of deviation = 20
= 900
13.
The shortest length of a mirror which a man can see his full length is half of the height of man.
14.
Focal length of the plane mirror is 
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VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
15.
When a plane mirror is rotated through an angle  , the reflected ray turns through an angle 2  , then the
size of the image is Remains the same
16.
An object is placed symmetrically between the plane mirrors inclined at angle of 300. Then the total number
of images formed is 11
n
  300
360
360
1 
 1  11 images

30
17.
The image formed by a pin hole camera is sharpest when the hole is very small.
18.
When a convergent beam of light falls incident on a plane mirror , the image formed is Erect and real.
19.
A man stands in a room with his eyes at the centre of the room. The height of the ceiling is H. The length of
the shortest plane mirror, fixed on the wall in front of the man, so that the man can see full image of the wall
behind him is
H
.
3
20.
The danger signals are made red because the red colour is scattered minimum.
1.
Angle of incidence = 150
Accodrding to laws of reflection i  r  150
angle of deviation = 180 - 20 = 180 – 2(15)0 = 1500
2.
Given snake is approaching a stationary plane mirror with a speed of 5m is then it observes that its image
is approaching it with a speed of. 10 m/s.
3.
From given diagram
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300
600
0
0
i = 30 , Given  = 60 , that angle is angle between reflected ray with plane mirror, from this angle of
reflection = 90– 60 = 300.
4.
i  40 0
W.K.T i  r  40 0
5.
Image of B= 7+2(2)=11 m
6.
Govem distance between the mirror and his image is 4m, and he moves 1m towards the mirror, the distance
between shyam and his image will be
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4 + 2 (1) = 6m.
7.
The sunlight consists of seven colour i.e, ‘VIBGYOR’ represents converging beam.
8.
Light rays from different sources come to merge at apoint is conveiging beam and it is shown in given
diagram.
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9.
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
  720
n
n
360
1

360
 1 4
72
10.
A plane mirror reflecting a ray of light rotated through an angle 100 about an axis through the point of
incidence in the plane of the mirror and perpendicular to plane of incidence then the reflected ray rotates
through an angle of 200.
11.
Given i = 300 angle of incidence = angle of reflection (according to laws of reflection).
0
i = r = 30
12.
You are standing in front of a plane mirror. which is approaching you with a speed of 10cm/sec. Then the
speed of the image approaching you will be 10 cm/s.
13.
Given   450
number of images =
360

=
14.
360
8
45
Given no. of images = 9 then 
n

360
1

360
 n  1  10

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  36
15.
r  300 ,
i  300
= 2  2(30)  600
angle of deviation = 180  2  1200
16.
A ray of light is incident on a plane mirror at 
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deviation produced by the mirror is 2  .
17.
The image formed in a plane, mirror is as far behind as the objects is in front of it. If we approach a mirror
with speed v the image approaches us with 2v.
18.
The between incident ray and veritical upward is 400, and that between reflected ray and normal to the
plane mirror is 700, that between normal and horizontal is 200 also as the angle between a pair of lines is
equal to the angle between their normal therefore, angle between vertical and plane mirror is 200, that
between mirror and horizontal is 700.
19.
Deviation after the reflection from first mirror is   1200
That after reflection from the 2nd mirror is also   1200
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VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
M1

  120 0
60
600
0
0
300
30
60 0
O
M2
Total deviation 120 = 2400.
Eye
h
20.
From the geometry of figure , h=60 m
BRAIN TWISTERS
1.
A plane mirror reflecting a ray of incident light is rotated through an angle  about an axis through the
point of incidence in the plane of the morror perpendicular to the plane of incidence. Then the reflected ray
rotates through an angle 2  and the incident ray is fixed .
2.
The angle according to laws of reflection incidence is equal to the angle of reflection, the angle of incidence
is equal to the angle of reflection. According to laws of reflection.
3.
The image is formed behind the mirror has the same size as the object, image is laterally inverted, if image
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is virtual it cannot be received on a scree, the image is as far behind the mirror as the object is in front of it
the Characteristics of an image formed by a plane mirror.
4.
The normal by an angle  , the reflected ray rotates by an angle 2  . If a plane mirror is rotated by an angle
are plane mirror is rotated by an angle 1e  for a fixed incident ray..
5.
The angle of deviation ‘d’ of light produced is 1800 – 2i
The maximum value of angle of deviation is 1800
6.
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7.
A. Reflection
-
Silcer
B. Lateral inversion
-
Plane mirror
C. Translucent
-
Waxed paper
D. Luminous object
-
Sun
a) Regular reflection
-
polished smooth surfaces
b) Irregular reflection
-
c) Angle of incidence
-
Angle between normal and incident ray
d) Angle of reflection
-
angle between normal and reflected ray
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rough surfaces
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CHAPTER SOLUTIONS - 2
File No.33/07/18/11/2014
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
8.
a) Virtual image
-
It is always erect ‘n’ reflections from two plane mirrors inclined
at an angle  is given by
b) Real image
-
It is always inverted due to reflection
c) The deviation produced by
-
n(180-  )
d) Formula for the angle of deviation 9.
a)  = 600
-
n=5
b)  = 400
-
n=9
c)  = 900
-
n=3
d)  = 1800
-
n=5
For  , no. of images =
10.
180-2i
360
-1

Light travels in a straight line as long as it is travel in a same medium and light appears to travel in straight
lines.
Both assertion and reason are correct and reason explains assertion.
11.
For a normal incidence. the angle of deviation is 1800. The angle between the incident ray extended and the
reflected ray is called the angle of deviation.
Both assertion and reason are correct and reason does not explains assertion.
12.
Light comes back into same medium from a boundary surface seperating two media is called reflection and
the surface from which reflection takes places is called reflector.
Both assertion and reason are correct and reason explains assertion.
13.
In reflection , angle of incidences is always equal to angle of reflection, if a ray incident along normal to the
mirror reflects its path.
14.
In case of a plane mirror , distance of object and its image is eqaul from any point on the mirror, Keeping
a point object fixed, if a plane mirror is moved
image will also move.
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Both assertion and reason and correct and reason explains assertion.
15.
W.K.T. no. of images n 
360

if   720 , n 
360
=5
72
So, both assertion reason are correct and reason explains assertion.
16.
n
360

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given n = 7,   ?
17.
n
360
360
360
1 
 n  1  
 450


8
n
360
1

180  2  180  120  60 0
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CHAPTER SOLUTIONS - 2
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18.
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
For given problem
900
diagram
r
60
from figure, i = 90 – 60 = 300.
19.
i  1200 ,
20.
180  2  180  120  60 0
1.
If a spherical mirror is immersed in a liquid. it’s focal length will remain unchanged.
2.
Mark correct option or options : The minimum distance between a real image and the real object in concave
mirror is zero.
3.
4.
r  1200 because of law of reflection.
H 1.8

 0.9 m
2
2
An unlit candle is 90 cm from a concave mirror of radius 60 cm. Where must a flame be put from the mirror
so that its image may be seen coming out of the candle 45 cm.
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5.
1
1 1
   =  n  1  f
f
u nu  n 
6.
Vimage  2Vmirror  Vobject
speed of image with respect to mirror V = Vimage  Vmirror
7.
Incident ray is fixed, if mirror rotated through angle 150, then reflected ray will rotated through an angle =
300.
8.
If the incident ray and the reflected ray from a mirror are mutually perpendicular toeach other then angle of
incidence is the angle made by incident ray with normal is zero.
i
r
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9.
Angle between an incident beam and surface upon which it is indent. If a ray is incident an a mirror at a
glancing angle ‘2g’.
Angle of deviation is also 2g.
10.
Given two plane mirrors are inclined at an angle of 900. If a ray of light is incident on one mirror at 600,
then angle of deviation is 1800.
11.
The maximum angle of deviation possible for a ray incident on a plane mirror is 1800
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CHAPTER SOLUTIONS - 2
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12.
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
Given angle between incident and reflected rays is 600.
If mirror is rotated to increase the angle of incidence by 30, then now the angle between incident and reflected
ray is 60 + 6 = 66°.
13.
A ray of light reflected from a mirror rotated through 200 ,
i.e, 2  = 200
   100 , angle of rotation of the mirror..
14.
A ray of light from a lamp fall normally on a vertical plane mirror at a distance of 1m and its im age is
formed on a scale placed just above the lamp. When the mirror is rotated through an angle  about a vertical
axis, the spot of light on the scale is displaced through 10cm. Find  is 1.9°.
15.
A boy of height 1.5m with his eye level at 1.4m stands before a plane mirror of length 0.75 fixed on the well.
The height of the lower edge of the mirror above the floor is 0.8 m. Then the boy can’t see his feet.
16.
Identical twins stand at equal distances on opposite sides of an opaque wall. What is the minimum size of
a window which must be cut in the wall, so that they can obtain a full view of each other is half the size of
the twins.
17.
In reflection, i  r , and a ray incident along normal to the mirror reflects its path.
Both are correct.
18.
W.K.T. D  360  2  240  360  2
 2 120
  600
n
19.
360
1
60
n=5
One third of the height of the wall
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500

Incident ray
100
I
r
Horizontal
40 0
Reflected ray
20.
ir 
100
 500
2
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[The angle between incident ray and reflected ray is (900+100=1000)]
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