Physics Numericals On Reflection Of Light Class 9
Reflection Of Light Class 9
Numericals :.jfif)
1 . A ray is incident on a plane mirror. Its reflected ray is perpendicular to the incident ray. Find the angle of incident.
Ans : ∠ i = ∠ r = x° ( Law of reflection)
∠i + ∠ r = 90° (given)
x° + x° = 90°
2x = 90°
x = 45°
∠ i = 45°
2. A man standing in front of a plane mirror finds his image at a distance 6 metre from himself. What is the distance of man from the mirror ?
Ans : Since image form in a plane mirror is as far behind the mirror as the object is in front of it i.e.
distance of object from mirror = distance of image from mirror
Distance of man from mirror = 6 / 2 = 3m
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 ?
Ans : Here distance of insect in front of a plane mirror = 1m
(a) distance of image from mirror = 1 m
distance of image in a plane mirror is as far behind the mirror as the object is in front of it.
(b) distance between insect and its image is equal to 1+ 1 = 2
4. An optician while testing the eyes of a patient keeps a chart of letter 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 ?
Ans : Initially the distance between letters and patient = 3m
Distance of patient from the plane mirror = 2 m
Distance of image form in the mirror = 2 m
( Distance of image in a plane mirror is as far behind the mirror as the object is in front of it )
Distance between chart and patient = 3 + 2 + 2 = 7 m
5. The radius of curvature of a convex mirror is 40 cm. Find its focal length.
Ans : Radius of curvature R = 40 cm
Focal length = R / 2
= 40 / 2
= 20 cm
6. The focal length of A1 kilometer is 10 cm. Find its radius of curvature.
Ans : Focal length f = 10 cm
R = 2f = 2 * 10
R = 20 cm
7. 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.
Ans : O = 2 cm, I = ?
u = - 20 cm
f = -12 cm
1/V = 1/ f - 1/u
= - 1/12 - (- 1 / 20 )
= - 1 / 12 + 1 / 20
= -1/12 + 1/20
= -5 +3 /60
= -2/60
= - 1/30
V = 30 cm
I/O = V/u
1/2 = -30 / -20 = 1.5
I = 2 * 1.5 = 3 cm
Nature of Image = Real, inverted, magnified
8. At what distance from a concave mirror of focal length 25 cm should an object be placed so that the size of image is equal to the size of object.
Ans : f= 25 cm, u = ?
Size of image is equal to size of object.
The object must be placed at C
i.e. at 2
f = 2 * 25 = 50 cm
9. A concave mirror forms a real image of an object placed in front of it at distance 30 cm, of size three times the size of object. Find (a) the focal length of mirror (b) position of image.
Ans : (a) :
Given
Distance from the object (u) = 30 cm
Distance from the image (v) = 90 cm
Focal length = ?
we know,
1 = 1 + 1
f u v
1 = 1 + 1
f -30 -90
(since, the formed image is real, u and v will be negative)
1 = -3-1
f -90
1 = -4
f -90
1 = 1
f 22.5
f = 22.5 cm
Thus the focal length = 22.5 cm
(b) Position of the image shall be 90 cm
Position from image = 3 times distance from object
Position from image = 3 * 30 = 90 cm
10) 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.
Ans : (a) :
Given
Distance from the object (u) = 5 cm
Distance from the image (v) = 10 cm
Focal length = ?
we know,
1 = 1 + 1
f u v
1 = 1 + 1
f -5 10
(since, the formed image is virtual, u will be negative)
1 = -2 +1
f -10
1 = -1
f -10
1 = 1
f 10
f = 10 cm
Thus the focal length = 10 cm
(b) Position of the image shall be 10 cm
Position from image = 2 times distance from object
Position from image = 2 * 5 = 10 cm
11) The image formed by a convex mirror is of size 1/3 of the size of object. How are u and v related ? Ans:
Given m = 1 / 3
since, magnification n is positive,
so it is virtual image.
m = - v
u
1 = -v
3 u
1u = -v
3
v = -1u
3
u = -3v
12) The erect image formed by a concave mirrors is of size double the size of object. How are u and v related ?
Ans : Given,
m = 2
Since , magnification m is positive,
It is virtual image and hence u is ' + ve '
m = - v
u
2 = - v
u
v = -2u
13) The magnification for a mirror is -3. How are u and v related ?
Ans : Given,
m = - 3
m = v
u
-3u = v
v = - 3u
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