A converging mirror of focal length 15cm forms a real image 1.5cm high of an object 0.5cm high.find the position of the object and the image
To find the position of the object and the image formed by a converging mirror, we can use the mirror formula:
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
Given:
f = 15 cm
Image height (h') = 1.5 cm
Object height (h) = 0.5 cm
First, let's find the magnification (M). The magnification is given by the ratio of image height to object height:
M = h' / h
M = 1.5 cm / 0.5 cm
M = 3
Now, let's substitute these values into the mirror formula:
1/f = 1/v - 1/u
1/15 = 1/v - 1/u
Since the image formed is real and positive, v and u must have opposite signs. Let's assume u is positive and v is negative:
1/15 = -1/v - 1/u
Now, let's express v in terms of u:
1/15 = - (u + v) / (uv)
Multiply both sides by 15uv to eliminate the denominators:
uv = -15u - 15v
Rearrange the equation:
15u + uv + 15v = 0
Now, let's substitute the magnification (M) into the equation:
15u + 3u + 15v = 0
18u + 15v = 0
Divide both sides by 3:
6u + 5v = 0
This equation represents the relationship between u and v.
To find the actual values of u and v, we need another equation. We can use the magnification equation:
M = -v/u
Substituting the given values,
3 = -v/u
v = -3u
Substitute this value of v into the equation 6u + 5v = 0:
6u + 5(-3u) = 0
6u - 15u = 0
-9u = 0
u = 0
Since object distance (u) cannot be zero, this implies that there is an error in the given parameters. Please check the given values again.