A concave spherical mirror has a radius of curvature 15 cm. A 1 cm tall bulb is placed in front of the mirror such that its image is formed 10 cm in front of the mirror
what is the focal length
position of object
magnification of object
height of object
Please help!!
I am truly stuck
Have you drawn a ray diagram?
then, used the mirror equation?
When an object is placed at a distance of 15cm from a concave mirror its image is formed at a 10 cm in front of the mirror. Find the focal length
To solve this problem, we will use the mirror formula for a concave spherical mirror:
1/f = 1/v - 1/u
Where:
- f is the focal length of the mirror
- v is the image distance
- u is the object distance
1. Finding the focal length (f):
For a concave mirror, the focal length is considered negative. In this case, since the image is formed in front of the mirror, we can assume that the object distance (u) is negative.
Given:
Radius of curvature (R) = -15 cm
Image distance (v) = -10 cm
Object height (h) = 1 cm
Since the object distance (u) is not given directly, we can use the mirror equation:
1/f = 1/v - 1/u
Substituting the given values:
1/f = 1/-10 - 1/u
Now, let's find the object distance (u) using the magnification equation:
magnification (m) = -v/u
Substituting the given values:
1 - 10/u = -1
Solving for u:
10/u = 2
u = 5 cm
Now we can substitute the value of u back into the mirror equation to find the focal length:
1/f = 1/-10 - 1/5
1/f = -1/10 - 2/10
1/f = -3/10
f = -10/3 cm ≈ -3.33 cm
So, the focal length of the concave mirror is approximately -3.33 cm.
2. Finding the position of the object:
The position of the object is given by the object distance (u), which we have already found to be 5 cm.
The object is placed 5 cm in front of the mirror.
3. Finding the magnification of the object:
The magnification (m) is given by the formula:
magnification (m) = -v/u
Substituting the given values:
m = -(-10)/5
m = 2
So, the magnification of the object is 2.
4. Finding the height of the object:
The height of the object is given as 1 cm.
Therefore, the height of the object is 1 cm.
To summarize:
- Focal length (f) ≈ -3.33 cm
- Position of the object (u) = 5 cm
- Magnification (m) = 2
- Height of the object (h) = 1 cm