Determine the magnification for an object placed 25.9 cm from a convex lens having a focal length of 12.2 cm.

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Question: Determine the magnification for an object placed 25.9 cm from a convex lens having a focal length of 12.2 cm.

Step 1: Identify the given values and choose a sign convention. We will use the common lens sign convention where the object distance u is taken as positive if the object is to the left of the lens. Here, u = 25.9 cm (positive by this convention), and the focal length f = 12.2 cm (positive for a convex lens).

Step 2: Use the lens equation to find the image position v. The lens formula is 1/f = 1/v + 1/u. Rearranging to solve for 1/v gives 1/v = 1/f − 1/u.

Step 3: Compute 1/f − 1/u with the given numbers:
- 1/f = 1/12.2 ≈ 0.08197
- 1/u = 1/25.9 ≈ 0.03861
- 1/v = 0.08197 − 0.03861 ≈ 0.04336

Step 4: Solve for v: v ≈ 1 / 0.04336 ≈ 23.0 cm (positive means the image forms on the opposite side of the lens, i.e., a real image).

Step 5: Determine the magnification m. The linear magnification is m = −v/u (the negative sign indicates the image is inverted relative to the object, given the chosen sign convention).

Step 6: Compute m with v ≈ 23.0 cm and u = 25.9 cm:
- m ≈ −(23.0 / 25.9) ≈ −0.89

Interpretation: The magnification is approximately −0.89, meaning the image is inverted and about 0.89 times the size of the object.

Notes on potential confusions:
- If you switch sign conventions (e.g., take u as negative for an object to the left), you would obtain a different intermediate value for v, but the final magnification using the standard m = −v/u with the conventional sign yields the same numerical magnitude and the negative sign indicating inversion.
- Always check the sign convention you’re using and apply the lens formula consistently to avoid sign errors.

Physics 1 Hon - Bowles - p06/p07 P10.22 - Lesson 6 Exit Ticket

Determine the magnification for an object placed 25.9 cm from a convex lens having a focal length of 12.2 cm.

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