An arrow 2.5cm high is placed at a distance of 25cm from a diverging mirror of focal length 20cm. Find the nature,position and size of the image formed.
To find the nature, position, and size of the image formed by a diverging mirror, we can use the mirror formula and the magnification formula.
The mirror formula is given by:
1/f = 1/v - 1/u
Where:
f = focal length of the mirror
v = image distance from the mirror
u = object distance from the mirror
The magnification formula is given by:
magnification (m) = -v/u
Given:
Height of the object (h) = 2.5 cm
Distance of the object (u) = -25 cm (negative sign indicates that the object is placed in front of the mirror)
We need to find:
Focal length (f), Image distance (v), and Size of the image (h')
Let's start by calculating the focal length:
Rewriting the mirror formula as a ratio:
1/f = (v - u) / (v * u)
Using the given values:
1/f = (v - (-25 cm)) / (v * (-25 cm))
Simplifying:
1/f = (v + 25 cm) / (-25v)
To find the position of the image (v), we need to solve this equation. Rearranging the formula gives us:
v = (-25v) / (v + 25 cm)
Simplifying further:
v + 25 cm = -25v
Bringing like terms on one side:
26v = -25 cm
Solving for v:
v = -25 cm / 26
v ≈ -0.96 cm (approximately)
The negative sign indicates that the image is formed on the same side as the object (i.e., it is a virtual image).
Now, let's calculate the size of the image (h'):
Using the magnification formula:
magnification (m) = h' / h
Rearranging the formula:
h' = m * h
Using the given height of the object (h) and the magnification formula:
h' = -v/u * h
h' = -(-0.96 cm) / (-25 cm) * 2.5 cm
h' ≈ 0.096 cm (approximately)
The positive value of the image height (h') indicates that the image is upright.
Therefore, the nature of the image formed by the diverging mirror is a virtual, upright image. It is formed approximately 0.96 cm behind the mirror and has a height of approximately 0.096 cm.