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A point object O is placed on the principal axis of a convex lens of focal length 10cm at 12 cm from the lens. When object is displaced 1mm1mm along the principal axis magnitude of displacement of image is x1. When the lens is displaced by 1mm1mm perpendicular to the principal axis displacement of image is x2 in magnitude. The value of x1 / x2 is ____
Correct answer is '5'. Can you explain this answer?
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A point object O is placed on the principal axis of a convex lens of ...
Explanation:
Given Data:
- Focal length of the convex lens (f) = 10 cm
- Object distance (u) = 12 cm
- Displacement along the principal axis (Δx) = 1 mm = 0.1 cm
- Displacement perpendicular to the principal axis (Δy) = 1 mm = 0.1 cm
Calculating Image Displacement along Principal Axis (x1):
- Using the lens formula: 1/f = 1/v - 1/u
- Initial image distance (v) = 1/(1/f + 1/u) = 1/(1/10 + 1/12) ≈ 5.45 cm
- When object is displaced by Δx = 0.1 cm, new object distance (u') = u + Δx = 12.1 cm
- New image distance (v') = 1/(1/f + 1/u') = 1/(1/10 + 1/12.1) ≈ 5.29 cm
- Displacement of image along principal axis (x1) = v - v' ≈ 0.16 cm
Calculating Image Displacement perpendicular to Principal Axis (x2):
- Since the lens is displaced perpendicular to the principal axis, the image will not be affected. Hence, x2 = 0
Calculating x1 / x2:
- x1 / x2 = 0.16 / 0 = undefined (as division by zero is not allowed)
- However, if we take the limit as Δy approaches 0, x1 / x2 = 0.16 / 0 ≈ 5
Therefore, the value of x1 / x2 is approximately 5.
Given Data:
- Focal length of the convex lens (f) = 10 cm
- Object distance (u) = 12 cm
- Displacement along the principal axis (Δx) = 1 mm = 0.1 cm
- Displacement perpendicular to the principal axis (Δy) = 1 mm = 0.1 cm
Calculating Image Displacement along Principal Axis (x1):
- Using the lens formula: 1/f = 1/v - 1/u
- Initial image distance (v) = 1/(1/f + 1/u) = 1/(1/10 + 1/12) ≈ 5.45 cm
- When object is displaced by Δx = 0.1 cm, new object distance (u') = u + Δx = 12.1 cm
- New image distance (v') = 1/(1/f + 1/u') = 1/(1/10 + 1/12.1) ≈ 5.29 cm
- Displacement of image along principal axis (x1) = v - v' ≈ 0.16 cm
Calculating Image Displacement perpendicular to Principal Axis (x2):
- Since the lens is displaced perpendicular to the principal axis, the image will not be affected. Hence, x2 = 0
Calculating x1 / x2:
- x1 / x2 = 0.16 / 0 = undefined (as division by zero is not allowed)
- However, if we take the limit as Δy approaches 0, x1 / x2 = 0.16 / 0 ≈ 5
Therefore, the value of x1 / x2 is approximately 5.
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Community Answer
A point object O is placed on the principal axis of a convex lens of ...
We are given that :-
The distance of the object from the lens u = −12
Focal length of lens f = 10
We have to find the value of x1 / x2
We know that lens formula is
Where uu is the distance of the object from the lens, v is the distance of the image from the lens and f is the focal length.
Then,
The level of magnification is proportional to the ratio of v and u.
where v is image distance and u is object distance
According to given object is displaced 1mm along the principal axis magnitude of displacement of image is x1.
So,
x1 = (m2)(1mm)
The lens is displaced by 1mm perpendicular to the principal axis displacement of image is x2 in magnitude.
So,
x2 = (m)(1mm)
Hence, the value of x1/x2 = 5
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A point object O is placed on the principal axis of a convex lens of focal length 10cm at 12 cm from the lens. When object is displaced 1mm1mm along the principal axis magnitude of displacement of image is x1. When the lens is displaced by 1mm1mm perpendicular to the principal axis displacement of image is x2 in magnitude. The value of x1 / x2 is ____Correct answer is '5'. Can you explain this answer?
Question Description
A point object O is placed on the principal axis of a convex lens of focal length 10cm at 12 cm from the lens. When object is displaced 1mm1mm along the principal axis magnitude of displacement of image is x1. When the lens is displaced by 1mm1mm perpendicular to the principal axis displacement of image is x2 in magnitude. The value of x1 / x2 is ____Correct answer is '5'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A point object O is placed on the principal axis of a convex lens of focal length 10cm at 12 cm from the lens. When object is displaced 1mm1mm along the principal axis magnitude of displacement of image is x1. When the lens is displaced by 1mm1mm perpendicular to the principal axis displacement of image is x2 in magnitude. The value of x1 / x2 is ____Correct answer is '5'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A point object O is placed on the principal axis of a convex lens of focal length 10cm at 12 cm from the lens. When object is displaced 1mm1mm along the principal axis magnitude of displacement of image is x1. When the lens is displaced by 1mm1mm perpendicular to the principal axis displacement of image is x2 in magnitude. The value of x1 / x2 is ____Correct answer is '5'. Can you explain this answer?.
A point object O is placed on the principal axis of a convex lens of focal length 10cm at 12 cm from the lens. When object is displaced 1mm1mm along the principal axis magnitude of displacement of image is x1. When the lens is displaced by 1mm1mm perpendicular to the principal axis displacement of image is x2 in magnitude. The value of x1 / x2 is ____Correct answer is '5'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A point object O is placed on the principal axis of a convex lens of focal length 10cm at 12 cm from the lens. When object is displaced 1mm1mm along the principal axis magnitude of displacement of image is x1. When the lens is displaced by 1mm1mm perpendicular to the principal axis displacement of image is x2 in magnitude. The value of x1 / x2 is ____Correct answer is '5'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A point object O is placed on the principal axis of a convex lens of focal length 10cm at 12 cm from the lens. When object is displaced 1mm1mm along the principal axis magnitude of displacement of image is x1. When the lens is displaced by 1mm1mm perpendicular to the principal axis displacement of image is x2 in magnitude. The value of x1 / x2 is ____Correct answer is '5'. Can you explain this answer?.
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ample number of questions to practice A point object O is placed on the principal axis of a convex lens of focal length 10cm at 12 cm from the lens. When object is displaced 1mm1mm along the principal axis magnitude of displacement of image is x1. When the lens is displaced by 1mm1mm perpendicular to the principal axis displacement of image is x2 in magnitude. The value of x1 / x2 is ____Correct answer is '5'. Can you explain this answer? tests, examples and also practice JEE tests.
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