A 1.2 cm long pin is placed perpendicular to the principal axis of a convex mirror of focal length 12cm. at a distance of 8cm from it.
a) find the location of the image .
b) find the height of the image.
c) Is the image erect or inverted.
Answers
Answered by
32
Answer:
a) 4.8 cm
b) 0.72 cm
c) erect
Explanation:
For a convex mirror,
- f = positive
- u = negative
- v = positive
Where :
- f = focal length
- u = object distance
- v = image distance
Here :
- f = + 12 cm
- u = -8 cm
- v = ?
Using mirror formula,
- 1/v + 1/u = 1/f
- 1/v = 1/f - 1/u
- 1/v = 1/12 + 1/8
- v = 5/24
- v = 4.8 cm
∴ a) Image is formed on the right of the mirror at a distance of 4.8 cm from it, since v is positive.
Magnification,
- m = hₑ/hₒ = -v/u
Where :
- hₒ = object distance
- hₑ = image distance
Substituting :
- m = -4.8/-8
- hₑ = hₒ × -v/u
- hₑ = 1.2 × 0.6
- hₑ = 0.72 cm
∴ b) Height of the image (hₑ) is 0.72 cm.
∴ c) As hₑ is positive, image is on the same side of the principle axis as the object, so the image is erect.
Answered by
0
Answer:
Distance of object from mirror u=−8cm
Focal length f=12cm
We know for mirror,
f
1
=
v
1
+
u
1
12
1
=
v
1
+
−8
1
Distance of image from lens v=4.8cm
Magnification ratio m=−
u
v
=−
−8
4.8
=0.6
Height of object h=1.2cm
Height of the image h
i
=m×h=0.6×1.2=0.72cm
Explanation:
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