An object is 16.0 cm to the left of a lens. The lens forms an 36.0 cm to the right of the lens. (b) If the object is 8.00 mm tall, how tall is the ? Is it erect or inverted?

Verified step by step guidance

1. Identify the given values: object distance (do) is -16.0 cm (negative as it is to the left of the lens), and image distance (di) is 36.0 cm (positive as it is to the right of the lens).
2. Use the lens formula to determine the type of lens and the focal length. The lens formula is \(\frac{1}{f} = \frac{1}{do} + \frac{1}{di}\).
3. Calculate the magnification (m) using the formula \(m = -\frac{di}{do}\). The negative sign indicates the nature of the image (inverted if negative, erect if positive).
4. Determine the height of the image (hi) using the magnification formula \(hi = m \times ho\), where ho is the height of the object (8.00 mm).
5. Analyze the sign of the magnification to conclude whether the image is erect or inverted. If the magnification is positive, the image is erect; if it is negative, the image is inverted.

Key Concepts

Lens Formula

The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens through the equation 1/f = 1/v - 1/u. This formula is essential for determining the position of the image formed by the lens based on the position of the object.

Magnification

Magnification (M) is the ratio of the height of the image (h') to the height of the object (h), expressed as M = h'/h. It also relates to the distances of the object and image from the lens, given by M = -v/u. This concept helps determine the size and orientation of the image.

Image Orientation

The orientation of the image formed by a lens can be determined by the sign of the magnification. If M is positive, the image is erect; if M is negative, the image is inverted. Understanding this helps in predicting how the image will appear relative to the object.