Question

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, and (h) on the same side of the lens as Oor on the opposite side. Fill in the missing information, including the value of m when only an inequality is given, where only a sign is missing, answer with the sign.

Short answer

1. The lens type is converging.
2. The focal distance is +20cm.
3. The object distance is +8.0cm.
4. The image distance is -13cm .
5. The lateral magnification is +1.7 .
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

Given data

• The focal distance, f =20cm
• The object distance, p=+8.0cm
• The lateral magnification, m>1.0.

Understanding the concept of properties of the lens

An object, when placed in front of a lens, produces an image. It could be real or virtual, magnified or diminished, inverted or not inverted. The characteristics of the image are decided by the type of lens used, the focal length of the lens, and the distance of the object from the lens.

Formulae:

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

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

a) Calculation of the lens type

The magnification m is positive, and m>1.0. So the image is magnified and non-inverted. Hence the image will be virtual. A convergent lens can form a magnified, virtual, and non-inverted image.

Hence, the lens used is convergent.

b) Calculation of the focal distance

As the lens used is convergent, the focal length should be taken as positive, so it is +20cm.

c) Calculation of the object distance

From the given table, the object distance is given as +8.0 cm.

d) Calculation of the image distance

The image distance can be calculated using the given data in equation (1) as follows:

Hence, the image distance is.

$$\begin{gathered} \frac{1}{i}=\frac{1}{f}-\frac{1}{p} \\ =\frac{1}{20}-\frac{1}{8.0} \\ =\frac{-3}{40} \\ i=\frac{-40}{3.0} \\ =-13.3cm \\ =-13cm \end{gathered}$$

Hence, the image distance is -13cm.

e) Calculation of the lateral magnification

The lateral magnification can be calculated using the given data in equation (ii) as follows:

$$\begin{gathered} m=-\frac{(-13.3)}{8.0} \\ =+1.67 \\ =+1.7i.e.,m>0 \end{gathered}$$

Hence, the value of the magnification is +1.7.

f) Calculation of the type of image

The value of image distance is negative.

Hence, the image is virtual (V).

g) Calculation if the image is inverted or not

The value of lateral magnification is positive.

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

h) Calculation of the position of the object

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

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