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If an object of 10cm height is placed at a distance of 36cm form a converting mirror of focal length 12cm.Find the position,nature and height of the image?
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Given:
- Height of the object (h) = 10 cm
- Distance of the object from the mirror (u) = -36 cm (negative sign indicates that the object is placed in front of the mirror)
- Focal length of the mirror (f) = 12 cm

To find:
- Position of the image (v)
- Nature of the image (real/virtual, erect/inverted)
- Height of the image (h')

Solution:

Step 1: Calculating the position of the image (v)
Using the mirror formula:

1/f = 1/v - 1/u

Substituting the given values:

1/12 = 1/v - 1/-36

Simplifying the equation:

1/v = 1/12 + 1/36

1/v = 3/36 + 1/36

1/v = 4/36

v = 36/4

v = 9 cm

Therefore, the position of the image (v) is 9 cm.

Step 2: Determining the nature of the image
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Step 3: Finding the height of the image (h')
Using the magnification formula:

magnification (m) = -v/u

Substituting the given values:

m = -9/-36

m = 1/4

Therefore, the magnification is 1/4.

The height of the image can be found using the formula:

h' = m * h

Substituting the calculated magnification and given object height:

h' = (1/4) * 10

h' = 2.5 cm

Therefore, the height of the image (h') is 2.5 cm.

Summary:
- The position of the image (v) is 9 cm.
- The nature of the image is virtual and erect.
- The height of the image (h') is 2.5 cm.
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If an object of 10cm height is placed at a distance of 36cm form a converting mirror of focal length 12cm.Find the position,nature and height of the image?
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