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An object 2cm tall is placed at 10cm from a converging lens of focal length 6cm. find the size , nature and position of the image?
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An object 2cm tall is placed at 10cm from a converging lens of focal l...
Object Details
- Height: 2 cm
- Distance from lens (u): -10 cm (the negative sign indicates the object is on the same side as the incoming light)
- Focal length (f): 6 cm (positive for converging lens)
Image Position Calculation
Using the lens formula:
1/f = 1/v - 1/u
Substituting the values:
1/6 = 1/v - 1/(-10)
1/6 = 1/v + 1/10
To find a common denominator (30):
5/30 = 1/v + 3/30
1/v = 5/30 - 3/30
1/v = 2/30
v = 15 cm
Thus, the image is formed at a distance of 15 cm from the lens on the opposite side.
Size of the Image
To find the size (height) of the image (h'):
Using the magnification formula:
magnification (m) = h'/h = v/u
Substituting known values:
m = 15 / -10 = -1.5
Now, height of the image:
h' = m * h
h' = -1.5 * 2 = -3 cm
Nature of the Image
- Height: 3 cm (the negative sign indicates the image is inverted)
- Nature: Real and inverted
Summary of Image Characteristics
- Position: 15 cm from the lens on the opposite side
- Size: 3 cm tall
- Nature: Real and inverted
- Height: 2 cm
- Distance from lens (u): -10 cm (the negative sign indicates the object is on the same side as the incoming light)
- Focal length (f): 6 cm (positive for converging lens)
Image Position Calculation
Using the lens formula:
1/f = 1/v - 1/u
Substituting the values:
1/6 = 1/v - 1/(-10)
1/6 = 1/v + 1/10
To find a common denominator (30):
5/30 = 1/v + 3/30
1/v = 5/30 - 3/30
1/v = 2/30
v = 15 cm
Thus, the image is formed at a distance of 15 cm from the lens on the opposite side.
Size of the Image
To find the size (height) of the image (h'):
Using the magnification formula:
magnification (m) = h'/h = v/u
Substituting known values:
m = 15 / -10 = -1.5
Now, height of the image:
h' = m * h
h' = -1.5 * 2 = -3 cm
Nature of the Image
- Height: 3 cm (the negative sign indicates the image is inverted)
- Nature: Real and inverted
Summary of Image Characteristics
- Position: 15 cm from the lens on the opposite side
- Size: 3 cm tall
- Nature: Real and inverted
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An object 2cm tall is placed at 10cm from a converging lens of focal length 6cm. find the size , nature and position of the image?
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An object 2cm tall is placed at 10cm from a converging lens of focal length 6cm. find the size , nature and position of the image? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about An object 2cm tall is placed at 10cm from a converging lens of focal length 6cm. find the size , nature and position of the image? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An object 2cm tall is placed at 10cm from a converging lens of focal length 6cm. find the size , nature and position of the image?.
An object 2cm tall is placed at 10cm from a converging lens of focal length 6cm. find the size , nature and position of the image? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about An object 2cm tall is placed at 10cm from a converging lens of focal length 6cm. find the size , nature and position of the image? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An object 2cm tall is placed at 10cm from a converging lens of focal length 6cm. find the size , nature and position of the image?.
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