When applying the thin lens formula (equation 21.3), a real image has a positive image distance, d_{i} < 0. A virtual image has a negative image distance, d_{i} < 0. A negative image distance means the image is on the same side of the lens as the object.

Why is there a minus sign in equation (21.4)? Look at the ray diagram for a convex lens on page 1226. The image is inverted, which corresponds to the minus sign in equation (21.4). A negative magnification means that the image is inverted, whereas a positive magnification is an upright image.

Concave lenses and the thin lens formula

How can you use the thin lens formula for a concave or diverging lens? A diverging lens has a negative curvature, so its focal length is negative. Converging lenses have positive curvature and positive focal lengths. The thin lens formula will predict the image location for both converging and diverging lenses if you make sure to use the appropriate sign for the focal length.

Virtual images and the thin lens formula

A magnifying glass with a focal length of 20 cm is held 15 cm above the page of a book.
(a) How far from the lens is the image located? (b) Is the image real or virtual?
(c) What is the magnification of this image? (d) Is the image upright or inverted?

Asked:

(a) image distance d_{i};
(b) whether image is real (d_{i} > 0) or virtual (d_{i} < 0);
(c) magnification m of the image;
(d) whether the image is upright (m > 0) or inverted (m < 0)

Given:

object distance d_{o} = 15 cm, focal length f = 20 cm

(a) Solve for 1/d_{i} in the thin lens formula and calculate:
$$\frac{1}{{d}_{i}}=\frac{1}{f}-\frac{1}{{d}_{o}}=\frac{1}{20\text{cm}}-\frac{1}{15\text{cm}}=(0.05-0.0667){\text{cm}}^{-1}=-0.0167{\text{cm}}^{-1}$$
Invert both sides of the equation to get d_{i} = 1/(−0.0167 cm^{−1}) = −60 cm.
(b) The image distance is negative, so this is a virtual image and it is therefore located on the same side as the object.
(c) Use the magnification equation:

(d) The magnification is positive, so the image is upright.

Answer:

(a) d_{i} = −60 cm.
(b) It is a virtual image.
(c) Magnification m = +4.
(d) It is an upright image.

Test your knowledge

When a 20-cm-tall object is placed 40 cm from a lens, an image forms 60 cm from the lens on the other side. What is the magnification?

−3/2

−2/3

2/3

1/3

The correct answer is a. The image is enlarged and inverted.
Asked: magnification m of this system
Given: object distance d_{o} = 40 cm; image distance d_{i} = 60 cm
Relationships:m = −d_{i} ⁄ d_{o} Solve:$$m=-\frac{{d}_{i}}{{d}_{o}}=-\frac{60\text{cm}}{40\text{cm}}=-\text{}\frac{3}{2}$$

Test your knowledge

Consider the same 20-cm-tall object placed 40 cm from a lens, with an image forming 60 cm from the lens on the other side. What is the height of the image?

−60 cm

−30 cm

−6.66 cm

13.33 cm

The correct answer is b.
Asked: image height h_{i} Given: object height h_{o} = 20 cm;
magnification m = −3/2 (previous answer)
Relationships:m = h_{i} ⁄ h_{o} Solve:
Rearrange and solve equation for h_{i}:
$${h}_{i}=m\times {h}_{o}=-\text{}{\scriptscriptstyle \frac{3}{2}}\times 20\text{cm}=-30\text{cm}$$