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An object is placed ata a distance of 15 froma diverging lens of focal length 6cm. Find ita nature n position?
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An object is placed ata a distance of 15 froma diverging lens of focal...
Analysis of Object Position and Nature in Relation to a Diverging Lens

Given Data:
- Object distance (u) = -15 cm
- Focal length (f) = -6 cm

Calculation of Image Position:
- Using the lens formula: 1/f = 1/v - 1/u
- Substituting the given values: 1/-6 = 1/v - 1/-15
- Solving for image distance (v) gives: v = -10 cm

Determination of Image Nature:
- As the image distance (v) is negative, the image formed is virtual.
- Since the image is formed on the same side as the object, it is erect.
- The nature of the image is diminished as it is smaller than the object.

Conclusion:
- The image formed by the diverging lens is virtual, erect, and diminished.
- The position of the image is at a distance of 10 cm on the same side as the object.
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An object is placed ata a distance of 15 froma diverging lens of focal length 6cm. Find ita nature n position?
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