Question

69 through 79 76, 78 75, 77 More lenses. Object Ostands on the central axis of a thin symmetric lens. For this situation, each problem in Table 34-8 refers to (a) the lens type, converging (C)or diverging (D), (b) the focal distance f, (c) the object distance p, (d) the image distance i, and (e) the lateral magnification m. (All distances are in centimeters.) It also refers to whether (f) the image is real(R) or virtual (V), (g) inverted (I)or non-inverted (NI)from O, and (h) on the same side of the lens as Oor on the opposite side. Fill in the missing information, including the value of mwhen only an inequality is given, where only a sign is missing, answer with the sign.

Short answer

1. The lens type is diverging
2. The focal distance is -10cm
3. The object distance is +10cm
4. The image distance is -5.0cm.
5. The lateral magnification is +0.50
6. The image is virtual (V)
7. The image is non-inverted (NI)
8. The image is on the same side of the lens as the object.

Step-by-step solution

The given data

1. The object distance, p=+10cm
2. The lateral magnification, m=0.50
3. The image is not inverted.

Understanding the concept of properties of the lens

Formula:

The lens formula, $\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$ (i)

The magnification formula of the lens,$m=-\frac{i}{p}$ (ii)

Calculation of the lens type

(a)

The image formed is smaller than the object as the magnification m<1.0. Also the image is not inverted. So the image will be virtual. A divergent lens always forms a reduced image, i.e., m<1.0 .

Hence, the type of lens is diverging.

Calculation of the focal distance

(b)

The image distance is given using the given data in equation (ii) as follows:

$$\begin{gathered} i=-mp \\ =-0.5\times 10 \\ =-5.0cm \end{gathered}$$

As the lens used is divergent, the focal distance should be taken as negative. Now, using the given data in equation (i), the focal length of the lens can b given using the data in equation (i) as follows:

$$\begin{gathered} \frac{1}{f}=\frac{1}{10}+\frac{1}{-5.0} \\ =\frac{-1}{10} \\ =-10cm \end{gathered}$$

Hence, the focal length is -10cm.

Calculation of the object distance

(c)

From the given data in table, the object distance is+10cm.

Calculation of the image distance

(d)

The image distance is given using the given data in equation (ii) as follows:

$$\begin{gathered} i=-mp \\ =-0.5\times 10 \\ =-5.0cm \end{gathered}$$$$\begin{gathered} i=-mp \\ =-0.5\times 10 \\ =-5.0cm \end{gathered}$$

The image distance is -5.0cm.

Calculation of the lateral magnification

(e)

From the given data in table, the lateral magnification is +0.50.

Calculation of the type of image

(f)

The value of image distance is negative.

Hence, the image is virtual (V)

Calculation if the image is inverted or not

(g)

The value of lateral magnification is positive.

Hence, the image is non-inverted (NI).

Calculation of the position of the object

(h)

From the above data, it is given that p<f.

Hence, the image is on the same side as object.