Question

In Fig. 34-26, stick figure O stands in front of a thin, symmetric lens that is mounted within the boxed region; the central axis through the lens is shown. The four stick figures${\mathit{I}}_{\mathbf{1}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{I}}_{\mathbf{4}}$suggest general locations and orientations for the images that might be produced by the lens. (The figures are only sketched in; neither their height nor their distance from the lens is drawn to scale.) (a) Which of the stick figures could not possibly represent images? Of the possible images, (b) which would be due to a converging lens, (c) which would be due to a diverging lens, (d) which would be virtual, and (e) which would involve negative magnification?

Short answer

(a) The sick figures that could not possibly represent images are$I_2\text{and}{I}_3$.

(b) The sick figures that represent a converging lens are$I_1\text{and}{I}_4$.

(c) The sick figure that represents a diverging lens is$I_1$.

(d) The sick figure that represents a virtual magnification is$I_1$.

(e)The sick figure that represents a negative magnification is $I_4$.

Step-by-step solution

Definition of a converging lens and diverging lens

The lens that allows the rays of light to focus on one point is termed a converging lens and the lens that allows the rays of light from one point to scatter at different points is termed a diverging lens.

(a) Determination of the sick figures that could not possibly represent images

The lens forms an image on the left of the lens as upright and on the right of the lens as inverted.

Thus, the sick figures that could not possibly represent images are $I_2\text{and}{I}_3$.

(b) Determination of the sick figures that represent a converging lens

The converging lens forms an image on the left of the lens as upright and on the right of the lens as inverted.

Thus, the sick figures that represent a converging lens are $I_1\text{and}{I}_4$.

(c) Determination of the sick figures that represent a diverging lens

The image due to a diverging lens is always virtual and is located between the object and the lens.

Thus, the sick figure that represents a diverging lens is $I_1$.

(d) Determination of the sick figure that represents a virtual magnification

From the figure, it can be observed that $I_1$is a virtual image due to the lens.

Thus, the sick figure that represents a virtual magnification is $I_1$.

(e) Determination of the sick figure that represents a negative magnification

It is known that when the image is inverted, there would be a negative magnification.

Thus, the sick figure that represents a negative magnification is $I_4$.