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An object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis. The relative velocity between object and its image will be :
- a)increasing
- b)decreasing
- c)first increases then decreases
- d)first decreases and then increases
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
An object approaches a fixed diverging lens with a constant velocity f...
as u increased Vrel decrease
here u is negetive so when it approach lens it will increase.
Most Upvoted Answer
An object approaches a fixed diverging lens with a constant velocity f...
Explanation:
When an object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis, the relative velocity between the object and its image will decrease.
Reasoning:
1. The lens formula for a diverging lens is given by:
1/f = (1/v) - (1/u)
where f is the focal length of the lens, v is the image distance, and u is the object distance.
2. As the object approaches the lens from infinity along the principal axis, the object distance u decreases.
3. Since the object is approaching the lens, the object distance u is negative.
4. As the object distance u decreases, the magnitude of the first term on the right-hand side of the lens formula [(1/v)] increases.
5. The image distance v is also negative for a diverging lens, indicating that the image is formed on the same side as the object.
6. As the object approaches the lens, the image distance v also decreases.
7. The relative velocity between the object and its image is given by the magnitude of the difference between their velocities:
Relative velocity = |v - u|
8. Since both v and u are negative, their difference (v - u) will be negative.
9. As the object approaches the lens, both v and u decrease, but the magnitude of the difference between them decreases.
10. Therefore, the relative velocity between the object and its image decreases as the object approaches the lens.
Hence, the correct answer is option 'B' - the relative velocity between the object and its image decreases.
When an object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis, the relative velocity between the object and its image will decrease.
Reasoning:
1. The lens formula for a diverging lens is given by:
1/f = (1/v) - (1/u)
where f is the focal length of the lens, v is the image distance, and u is the object distance.
2. As the object approaches the lens from infinity along the principal axis, the object distance u decreases.
3. Since the object is approaching the lens, the object distance u is negative.
4. As the object distance u decreases, the magnitude of the first term on the right-hand side of the lens formula [(1/v)] increases.
5. The image distance v is also negative for a diverging lens, indicating that the image is formed on the same side as the object.
6. As the object approaches the lens, the image distance v also decreases.
7. The relative velocity between the object and its image is given by the magnitude of the difference between their velocities:
Relative velocity = |v - u|
8. Since both v and u are negative, their difference (v - u) will be negative.
9. As the object approaches the lens, both v and u decrease, but the magnitude of the difference between them decreases.
10. Therefore, the relative velocity between the object and its image decreases as the object approaches the lens.
Hence, the correct answer is option 'B' - the relative velocity between the object and its image decreases.
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An object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis. The relative velocity between object and its image will be :a)increasingb)decreasingc)first increases then decreases d)first decreases and then increasesCorrect answer is option 'B'. Can you explain this answer?
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