A converging lens (f=12.0 cm) is held 8.00 cm in front of a newspaper that has a print size with

step 1 Solving party of this problem, using the thin lens I can write 1 by f is equal to 1 by DI plus 1

A converging lens (f=12.0 cm) is held 8.00 cm in front of a...

A converging lens $(f=12.0 \mathrm{cm})$ is held $8.00 \mathrm{cm}$ in front of a newspaper that has a print size with a height of $2.00 \mathrm{mm}$. Find (a) the image distance (in $\mathrm{cm}$ ) and (b) the height (in $\mathrm{mm}$ ) of the magnified print.

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Key Concepts

Virtual Image Formation

When an object is positioned within the focal length of a converging lens, the resulting image is virtual. This means that the image forms on the same side as the object, cannot be projected onto a screen, and is typically larger and upright compared to the original object.

Sign Conventions

Sign conventions in optics are essential for correctly applying the thin lens equation and determining magnification. They dictate the assignment of positive or negative values to distances and heights based on the orientation and position of the object and image, ensuring consistent and accurate predictions in optical calculations.

Thin Lens Equation

The thin lens equation, given by 1/f = 1/do + 1/di, relates the focal length of a lens to the distances of the object (do) and the image (di) from the lens. It serves as a fundamental tool in geometrical optics for predicting where an image will form based on the placement of an object relative to the lens.

Magnification

Magnification is defined as the ratio of the image height to the object height and is also expressed as m = -di/do. This concept explains how the size of the image compares to the size of the object and indicates whether the image is inverted (negative magnification) or upright (positive magnification).

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