Question

“The value of magnification ‘m’ for a lens is −2.” Using the new Cartesian Sign Convention and considering that an object is placed at a distance of 20 cm from the optical centre of this lens, state:

1. the nature of the image formed;
2. size of the image compared to the size of the object;
3. position of the image, and
4. sign of the height of the image.

Answer 1

Using the new Cartesian sign convention, we take distances measured from the optical centre in the direction of the incident light (from left to right) as positive. Thus, the object placed on the left of the lens is at:

u = +20 cm

For a thin lens under this convention, the linear magnification is given by

`m = v/u`

so the image distance is

v = m.u = (−2) (20 cm) = −40 cm

1. The nature of the image formed:

The negative value of v (i.e. v = −40 cm) indicates that the image is formed on the side opposite to the incident light direction. In this convention, that means the image is real.

The magnification m = −2 is negative, which tells us that the image is inverted relative to the object.

2. Size of the image compared to the size of the object:

The magnitude of the magnification is ∣m∣ = 2.

Therefore, the image is twice as large as the object.

3. Position of the Image:

The image is located 40 cm from the optical centre, on the opposite side to the object (to the right of the lens).

4. Sign of the Height of the Image:

Since m = `(h_i)/(h_o)` = −2, if we take the object’s height h_o​ as positive (by convention), then the image height h_i​ must be negative. A negative image height indicates that the image is inverted relative to the object.