An image 5cm in height is placed at a distance of 15cm in front of a convex mirror of radius of curvature of 40cm.Find the position of the image and nature of the image.
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Answered by
1
Answer:
Radius of curvature (R) = 30 cm
f = R/2 = 30/2 = 15 cm
u = -20 cm,
h= 5 cm.
1/v +1/u = 1/f
1/v = 1/f – 1/u
1/v = 1/15 – 1/-20
v = 8. 57 cm
Image is virtual and erect and formed behind the mirror.
m = -v/u
m = hi/ho
hi/5= 8.57/20
= 0.428
hi = 0.428 x 5 = 2.14cm
Position of Image: Behind the mirror.
Nature of Image: Virtual and Erect.
Size of the Image: Diminished.
Answered by
0
Formula:
v
1
+
u
1
=
f
1
Given,
v
1
+
(−20)
1
=
15
1
v
1
=
60
7
⇒v=
7
60
v=8.57≈8.6cm
The image is formed behind the mirror at a distance of 8.6 cm
m=
h
1
h
2
=−
u
v
→
5
h
2
=−
20
8.57
∴h
2
=2.175 cm
Thus the image is 2.175 cm high, virtual and erect.
Hence, this is the answer.
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