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As an object is moved from the surface o...

As an object is moved from the surface of a thin converging lens to a focal point, over what range does the image distance very?

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To solve the problem of how the image distance varies as an object is moved from the surface of a thin converging lens to its focal point, we can follow these steps: ### Step 1: Understand the Lens Formula The lens formula is given by: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] where: - \( v \) is the image distance, - \( u \) is the object distance (with a negative sign for real objects), - \( f \) is the focal length of the lens. ### Step 2: Case 1 - Object at the Surface of the Lens When the object is at the surface of the lens, we can consider the object distance \( u \) to be 0 (since the surface of the lens is our reference point). Substituting \( u = 0 \) into the lens formula: \[ \frac{1}{v} - \frac{1}{0} = \frac{1}{f} \] This simplifies to: \[ \frac{1}{v} = \frac{1}{f} \] Thus, we find: \[ v = 0 \] This means that the image is formed at the lens itself when the object is at the surface. ### Step 3: Case 2 - Object at the Focal Point Next, we consider the case when the object is at the focal point of the lens. Here, the object distance \( u \) is equal to \(-f\) (the negative sign indicates that the object is on the same side as the incoming light). Substituting \( u = -f \) into the lens formula: \[ \frac{1}{v} - \frac{1}{-f} = \frac{1}{f} \] This simplifies to: \[ \frac{1}{v} + \frac{1}{f} = \frac{1}{f} \] Rearranging gives: \[ \frac{1}{v} = 0 \] Thus, we find: \[ v = \infty \] This means that when the object is at the focal point, the image is formed at infinity. ### Step 4: Conclusion From the two cases analyzed: - When the object is at the surface of the lens, the image distance \( v \) is 0. - When the object is at the focal point, the image distance \( v \) approaches infinity. Thus, we conclude that as the object moves from the surface of the lens to the focal point, the image distance varies from \( 0 \) to \( \infty \). ### Final Answer The image distance varies from \( 0 \) to \( \infty \). ---

To solve the problem of how the image distance varies as an object is moved from the surface of a thin converging lens to its focal point, we can follow these steps: ### Step 1: Understand the Lens Formula The lens formula is given by: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] where: ...
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