A 4.5cm needle is placed 12 centimetre away from a convex mirror of focal length 15 CM. give the location of image and magnification.
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Hi..
Height of the needle (h1) = 4.5 cm
Object distance( u) = – 12 cm
Focal length of the convex mirror (f)= 15 cm
Image distance = v
The value of v can be obtained using the mirror formula:
1/f = 1/v - 1/u
1/u = 1/v - 1/f
= 1/-25 - 1/5
= -5 -1/25
= -6/25
u = -25/6 = -4.167 cm
1/v = 1/f - 1/u
1/f = 1/u + 1/v
= 1/15 + 1/12
= 4 +5/60
= 9/60
f = 60/9 = 6.7 cm
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is
m = h2/h1 = -v/u
h2 = -v x h1 /u
= -6.7 x 4.5/ -12 = 2.5 cm
Hence Hence, magnification of the image, m = h2/h1 = 2.5/4.5 = 0.56
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce
Hope this helps u!!
Height of the needle (h1) = 4.5 cm
Object distance( u) = – 12 cm
Focal length of the convex mirror (f)= 15 cm
Image distance = v
The value of v can be obtained using the mirror formula:
1/f = 1/v - 1/u
1/u = 1/v - 1/f
= 1/-25 - 1/5
= -5 -1/25
= -6/25
u = -25/6 = -4.167 cm
1/v = 1/f - 1/u
1/f = 1/u + 1/v
= 1/15 + 1/12
= 4 +5/60
= 9/60
f = 60/9 = 6.7 cm
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is
m = h2/h1 = -v/u
h2 = -v x h1 /u
= -6.7 x 4.5/ -12 = 2.5 cm
Hence Hence, magnification of the image, m = h2/h1 = 2.5/4.5 = 0.56
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce
Hope this helps u!!
rahul648:
u can recheck it
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