An object whose height is 3.5 cm is at a distance of 8.5 cm from a spherical concave mirror. Its image is real and has a height of 11.2 cm. Calculate the radius of curvature of the mirror.
How far from the mirror is it necessary to place the above object in order to have a virtual image with a height of 11.2 cm?
To calculate the radius of curvature of the mirror, we can use the mirror formula:
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.
Given:
- h1 = 3.5 cm (object height),
- v1 = -11.2 cm (image height, negative sign indicates real image),
- u1 = -8.5 cm (object distance, negative sign indicates object is in front of the mirror).
Let's calculate the radius of curvature (f) using the given values:
1/f = 1/v1 - 1/u1
1/f = 1/(-11.2) - 1/(-8.5)
1/f ≈ -0.0893 + 0.1176
1/f ≈ 0.0283
f ≈ 35.26 cm
Therefore, the radius of curvature (R) is twice the focal length (f):
R = 2f ≈ 2 × 35.26 ≈ 70.52 cm
Now, let's calculate the object distance (u2) required to form a virtual image with a height of 11.2 cm:
Using the mirror formula again:
1/f = 1/v2 - 1/u2
Given:
- h2 = 11.2 cm (image height),
- v2 = infinity (virtual image is formed at infinity).
Since the virtual image is formed at infinity, we can assume that 1/v2 ≈ 0:
1/f ≈ 0 - 1/u2
1/f ≈ -1/u2
u2 ≈ -f
u2 ≈ -35.26 cm
Therefore, the object needs to be placed approximately 35.26 cm in front of the mirror to form a virtual image with a height of 11.2 cm.
To find the radius of curvature of the mirror, we can use the mirror formula:
1/f = 1/v - 1/u
Where:
f is the focal length of the mirror,
v is the image distance from the mirror, and
u is the object distance from the mirror.
Given:
Object distance (u) = -8.5 cm (negative because it is in front of the mirror),
Image distance (v) = -(-8.5 cm) = 8.5 cm (negative sign indicates that it is a real image),
Image height (h') = 11.2 cm.
Using the formula for magnification:
Magnification (m) = h'/h
where h is the object height.
Given:
Object height (h) = 3.5 cm,
Image height (h') = 11.2 cm.
Solving for magnification:
m = h'/h
m = 11.2 cm / 3.5 cm
m ≈ 3.2
Now, using the mirror formula:
1/f = 1/v - 1/u
1/f = 1/8.5 cm - 1/(-8.5 cm)
1/f = 2/8.5 cm
f ≈ 4.25 cm
Therefore, the radius of curvature of the mirror is approximately 4.25 cm.
To find the object distance required for a virtual image with a height of 11.2 cm, we can use the magnification formula:
m = -v/u (negative sign indicates a virtual image)
Given:
Magnification (m) = h'/h = 11.2 cm / 3.5 cm ≈ 3.2
Image height (h') = 11.2 cm
Solving for the object distance (u):
m = -v/u
3.2 = -11.2 cm / u
u ≈ -3.5 cm
Therefore, the object needs to be placed approximately 3.5 cm in front of the mirror to have a virtual image with a height of 11.2 cm.