91影视

1. Object distance,$p=+5.0cm$
2. The focal length of the lens, $f=+10cm$

Here, we need to use the concept of image formation by the thin lens. We can use equation 34.9 to solve for the image distance. The magnification of the lens can be calculated using equation 34.7. By using the values of image distance and magnification, and comparing the value of object distance and focal length we can determine whether the image is real or virtual, whether it is inverted or non-inverted, and whether it is on the same side as the object or on the opposite side.

(a)

The value of the focal length is positive.

Hence, the lens is a converging lens.

(b)

From the given data, the value of focal length is given as:$f=+10cm$.

Hence, the focal distance is $+10cm$.

(c)

From the given data, the value of object distance is given as:$p=+5.0cm$

Hence, the object distance is $+5.0cm$.

(d)

Now, using the above data in equation (i), we can get the image distance as follows:

$$\begin{gathered} \frac{1}{i}=\frac{1}{10\text{cm}}-\frac{1}{5.0\text{cm}} \\ =-0.1\text{cm} \\ i=-10\text{cm} \end{gathered}$$

Hence, the image distance is $-10cm$.

(e)

Using the given data in equation (ii), we can get the lateral magnification of the lens as follows:

$$\begin{gathered} m=-\frac{-10}{5.0} \\ =+2.0 \end{gathered}$$

Hence, the value of magnification is $+2.0$.

(f)

The value of image distance is negative.

Hence, the image is virtual (V).

(g)

The value of lateral magnification is positive.

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

(h)

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

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