TNPL PUBLIC SCHOOL (Senior Secondary), KAGITHAPURAM
Light - Reflection
PHYSICS
Class : 10
Roll No. : Time -
Date : MM - 38
1. The angle between incident ray and reflected ray is 60°. What is the angle of incidence? 1
Ans : Since
2. Name the mirror that can be used to check theft in shops. 1
Ans : Convex mirror.
3. What is the position of the object placed on the side of reflecting surface of a concave mirror of 1
focal length 15 cm if the image is formed at the distance of 30 cm from the mirror?
Ans : 30 cm
4. The image of an object formed by a mirror is real, inverted and is of magnification –1. If the 3
image is at a distance of 40 cm from the mirror, where is the object placed? Where would the image
be if the object is moved 20 cm towards the mirror? State reason and also draw ray diagram for the
new position of the object to justify your answer.
Ans : Given: Magnification of spherical mirror = –1, Image distance, v = –40 cm
Magnification, m =
u=
Therefore, the object is placed at a distance of 40 cm in front of the spherical mirror.
Case I: when u = – 40 cm and v = – 40 cm,
Using mirror formula, we get
Hence the focal length of the mirror is 20 cm, and the negative focal length shows that it
is a concave mirror.
The new position of the object when it moves 20 cm towards the concave mirror, u′ = –
(40 – 20) = –20 cm.
Case II: u' = –20 cm, f = – 20 cm, v′ = ?
From mirror formula,
Thus, the image is formed at infinity.
Hence when the object is moved 20 cm towards the mirror, a real, inverted and highly
enlarged
image is formed at infinity.
5. A spherical mirror produces an image of magnification –1 on a screen placed at a distance of 50 3
cm from the mirror.
(a) Write the type of mirror.
(b) Find the distance of the image from the object.
(c) What is the focal length of the mirror?
(d) Draw the ray diagram to show the image formation in this case.
Ans : (a) As magnification is negative, the image formed is real. Hence, it is a concave mirror.
(b)
u = v = – 50 cm
Distance of the image from the object = |u| + |v| = 100 cm
(c) Using mirror formula,
f = –25 cm
(d)
6. A student wants to obtain an erect image of an object using a concave mirror of 12 cm focal 3
length. What should be the range of distance of the object from the mirror? State the nature and
size of the image he is likely to observe. Draw a ray diagram to justify your answer.
Ans : If a student wants to obtain an erect image of an object using a concave mirror of 12 cm
focal length, he should keep the object between the pole and the focus of the mirror,
therefore, a virtual, erect and enlarged image will be formed behind the mirror.
7. For the given data showing the focal lengths of three concave mirrors A, B and C, and the 3
respective distances of different objects from these mirrors.
Object distance Focal length
S.No.
(cm) (cm)
A 45 20
B 30 15
C 20 30
Answer the following questions:
(i) In the given position of object from the mirrors, which mirror will form a diminished image of the
object. Draw a ray diagram for image formation by this mirror.
(ii) Which mirror can be conveniently used as a make-up mirror? Draw a ray diagram to illustrate
this function.
Ans : (i) Concave mirror A will form the diminished image of the object as the object is placed
beyond the centre of curvature (> 2f ) of the mirror.
(ii) Concave mirror 'C' can be used as a make-up mirror as the object distance is less
than the focal length of concave mirror, i.e. when the object is placed between the focus
'F' and the pole 'P' of the concave mirror, a virtual, erect and enlarged image is formed.
8. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the 3
location of the image and magnification. Describe what happens as the needle is moved farther
from the mirror?
Ans : Given: ho = + 4.5 cm, u = – 12 cm, f = + 15 cm
From mirror equation,
Also, magnification, m =
A virtual and diminished image is formed by a convex mirror at a distance of 6.7 cm
behind the mirror and the size of image is reduced to 2.5 cm.
As the needle is moved farther from the mirror, the image moves towards the focus and
will gradually reduce in size further.
9. List the new Cartesian sign convention for reflection of light by spherical mirrors. Draw a 5
diagram and apply these conventions for calculating the focal length and nature of a spherical
mirror which forms a 1/3 times magnified virtual image of an object placed 18 cm in front of it.
Ans : New Cartesian sign convention for reflection of light by spherical mirrors:
(i) The object is always placed to the left of the mirror.
(ii) All the distances parallel to the principal axis are always measured from the pole of
the spherical mirror.
(iii) All the distances measured along the direction of incident light, i.e. along +ve x-axis,
the corresponding quantities are considered to be positive.
(iv) All the distances measured opposite to the direction of incident light, i.e. along –ve x-
axis, the corresponding quantities are taken as negative.
(v) The distances measured in upward direction, i.e. perpendicular to and above the
principal axis along +ve y-axis, are taken as positive.
(vi) The distances measured in the downward direction, along –ve y-axis, i.e.
perpendicular to and below the principal axis, are taken as negative.
According to question, for a virtual image
m=
But, m =
Using mirror formula
f = + 9 cm
So, the focal length of the given spherical mirror is 9 cm. The Positive sign shows the
given mirror is convex in nature.
10. It is desired to obtain an erect image of an object, using concave mirror of focal length of 12 cm. 5
(i) What should be the range of distance of an object placed in front of the mirror?
(ii) Will the image be smaller or larger than the object? Draw ray diagram to show the formation of
image in this case.
(iii) Where will the image of this object be, if it is placed 24 cm in front of the mirror? Draw ray
diagram for this situation also to justify your answer.
Show the positions of pole, principal focus and the centre of curvature in the above ray diagrams.
Ans : (i) To obtain an erect image of an object, using concave mirror of focal length of 12 cm,
the object should be placed at a distance less than its focal length, i.e. 12 cm.
(ii) The obtained image will be larger than the object. The formation of image in this
case, is as shown in the adjoining figure.
(iii) If the object is placed at C( = 24 cm), in front of the concave mirror of focal length of
12 cm, then a real, inverted and same size image as that of the object is formed at C.
This situation is as shown in the adjoining figure.
11. Suppose you have three concave mirrors A, B and C of focal lengths 10 cm, 15 cm and 20 cm. 5
For each concave mirror you perform the experiment of image formation for three values of object
distances of 10 cm, 20 cm and 30 cm. Giving reason answer the following:
(a) For the three object distances, identify the mirror/mirrors which will form an image of
magnification –1.
(b) Out of the three mirrors identify the mirror which would be preferred to be used for shaving
purposes/make-up.
(c) For the mirror B draw ray diagram for image formation for object distances 10 cm and 20 cm.
Ans : (a) A real, inverted and same size image as that of the object formed by the concave
mirror will form an image of magnification –1. It is possible only when the object is
placed at C(R = 2f).
Hence for the object distances of 20 cm and 30 cm, the concave mirrors ‘A’ and ‘B’ will
form the real, inverted and same size image as that of the object. Therefore, the concave
mirrors ‘A’ and ‘B’ will form an image of magnification –1.
(b) The concave mirror ‘C’ of focal length 20 cm will be preferred to be used for shaving
purposes/ make-up. This is because when we bring our face within its focal length, it
forms a virtual, erect and enlarged image of our face.
(c) Ray diagram for image formation by mirror B
12. A student has focussed the image of a candle flame on a white screen using a concave mirror. 5
The situation is as given below:
Length of the flame = 1.5 cm
Focal length of the mirror = 12 cm
Distance of flame from the mirror = 18 cm
If the flame is perpendicular to the principal axis of the mirror, then calculate the following:
(a) Distance of the image from the mirror
(b) Length of the image.
If the distance between the mirror and the flame is reduced to 10 cm, then what would be observed
on the screen? Draw ray diagram to justify your answer for this situation.
Ans : Given: ho = + 1.5 cm, f = – 12 cm, u = – 18 cm
(a) For a concave mirror, using mirror formula
, we get
or
or v = –36 cm
So, the distance of the image from the mirror is 36 cm, the negative sign indicates that
the image is formed on the same side of the object.
(b) Using the formula m =
or hi =
So, the length of the image is 3.0 cm.
If the distance between the mirror and the flame is reduced to 10 cm, no image is formed
on the screen as the object lies between the focus and the pole of the mirror. So, a virtual
image behind the mirror is obtained as shown in the adjoining figure.
