Draw a ray diagram in each of the following cases to show the formation of image, when the object is placed:

(i) between optical centre and principal focus of a convex lens.

(ii) anywhere in front of a concave lens.

(iii) at 2F of a convex lens.

State the signs and values of magnifications in the above mentioned cases (i) and (ii).

To find the position and size of the image formed by the concave mirror, we use the mirror formula and magnification formula.

(i) Position of the Screen

Given:

• Size of the object, $h = +4.0\text{ cm}$
• Object distance, $u = -25.0\text{ cm}$ (by sign convention)
• Focal length of the concave mirror, $f = -15.0\text{ cm}$ (by sign convention)

Using the mirror formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Substituting the values:

$$\frac{1}{v} + \frac{1}{-25.0} = \frac{1}{-15.0}$$

$$\frac{1}{v} = -\frac{1}{15.0} + \frac{1}{25.0}$$

$$\frac{1}{v} = \frac{-5 + 3}{75.0} = -\frac{2}{75.0}$$

$$v = -37.5\text{ cm}$$

Thus, the screen should be placed at a distance of $37.5\text{ cm}$ in front of the mirror to obtain a sharp image.

(ii) Size of the Image

Using the magnification formula:

$$m = \frac{h'}{h} = -\frac{v}{u}$$

Substituting the values:

$$\frac{h'}{4.0} = -\frac{-37.5}{-25.0}$$

$$\frac{h'}{4.0} = -1.5$$

$$h' = -1.5 \times 4.0 = -6.0\text{ cm}$$

The negative sign indicates that the image is real and inverted. The size of the image is $6.0\text{ cm}$.

(iii) Ray Diagram

Since the object is placed at $25.0\text{ cm}$ (which is between the focus $F$ at $15.0\text{ cm}$ and the centre of curvature $C$ at $30.0\text{ cm}$), the image is formed beyond $C$ (at $37.5\text{ cm}$), and is real, inverted, and magnified.