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Answer :
a) The position of the image is approximately 5.44 cm.
the lateral amplification is approximately -0.32.
b) The position of the object is approximately 0.89 cm.
the lateral amplification is approximately 1.12.
a) Using the thin lens formula, which states that
1/f = 1/v - 1/u,
where f is the focal length of the lens, v is the position of the image, and u is the position of the object, we can solve for the unknown values.
Given:
Focal length (f) = 8.00 cm
Position of the object (u) = 17 cm
1/8.00 = 1/v - 1/17
Rearranging the equation, we have:
1/v = 1/8.00 + 1/17
1/v = (17 + 8.00)/(8.00 * 17)
1/v = 25.00/136.00
v = 136.00/25.00
v ≈ 5.44 cm
Therefore, the position of the image is approximately 5.44 cm.
To calculate the lateral amplification (m), we use the formula:
m = -v/u
m = -5.44/17
m ≈ -0.32
Therefore, the lateral amplification is approximately -0.32.
b) Given:
Position of the object (u) = ?
Position of the image (v) = -1.0 cm
Focal length (f) = 8.00 cm
Using the thin lens formula:
1/f = 1/v - 1/u
1/8.00 = 1/(-1.0) - 1/u
Rearranging the equation, we have:
1/u = 1/8.00 + 1/1.0
1/u = (1.0 + 8.00)/(8.00 * 1.0)
1/u = 9.00/8.00
u = 8.00/9.00
u ≈ 0.89 cm
Therefore, the position of the object is approximately 0.89 cm.
To calculate the lateral amplification (m), we use the formula:
m = -v/u
m = -(-1.0)/0.89
m ≈ 1.12
Therefore, the lateral amplification is approximately 1.12.
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