Hi,

Position of the image:

\[

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

\]

Given: \( f = 300 \text{ cm} \), \( u = -100 \text{ cm} \)

\[

\frac{1}{300} = \frac{1}{v} + \frac{1}{-100}

\]

\[

\frac{1}{v} = \frac{1}{300} + \frac{1}{100} = \frac{4}{300}

\]

\[

v = 75 \text{ cm}

\]

Nature of the image: Virtual, erect, and diminished (as always for convex mirrors).

Magnification:

\[

m = \frac{v}{u}

\]

Given: \( v = 75 \text{ cm} \), \( u = -100 \text{ cm} \)

\[

m = \frac{75}{-100} = -0.75

\]

As per given information,

f = 300 cm

u = -100 cm (negative sign because the object is in front of the mirror)

Using the Mirror Formula,

1/f = 1/v + 1/u

Substituting the values:

1/300 = 1/v + 1/(-100)

1/v = 1/300 + 1/100

1/v = (1 + 3) / 300

1/v = 4/300

v = 300/4

v = 75 cm

The image is formed 75 cm behind the mirror and the image is virtual, erect, and diminished.

Magnification (m) = -v/u

m = -75 / (-100)

m = 0.75

The magnification is 0.75, meaning the image is 0.75 times the size of the actual object.