91Ó°ÊÓ

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

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, $m=-\frac{i}{p}$

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.

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

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

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.

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.

The value of image distance is negative.

Hence, the image is virtual (V).

The value of lateral magnification is positive.

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

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

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