A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size.(u=-18.3cm, hi= 8.8 cm)
Answers
Hi,
Answer:
The image distance v = -18.3 cm and the image size hi = 8.8 cm.
Explanation:
Object height, ho = 4 cm
Object distance, u = 8.3 cm
Focal length, f = 15.2 cm
Let the image distance be denoted as “v” cm and the height of object and image be denoted as “ho” & “hi” respectively.
Using the lens formula for the given concave mirror, we get
1/f = 1/v + 1/u
⇒ 1/v = 1/f – 1/u
⇒ 1/v = 1/15.2 – 1/8.3
⇒ 1/v = 0.0657 – 0.1204 = - 0.0547
⇒ v = 1/0.0547
⇒ v = - 18.28 cm ≈ 18.3 cm
The magnification equation is defined as the ratio of the image distance (v) and the object distance (u) to the ratio of the image height (hi) and the object height (ho). The magnification equation is stated as follows:
M = [hi / ho ] = - [v / u]
∴ [hi / ho ] = - [v / u]
⇒ [hi / 4] = - [(-18.3) / 8.3]
⇒ hi = [18.3 * 4] / 8.3 = 8.8 cm
Hope this helpful!!!!!!
Answer:
Given
h=4cm
u= -8.3cm
f= -15.2cm(for concave mirror)
By mirror formula
1/v + 1/u = 1/f
1/v = 1/f - 1/u
1/v = 1/(-15.2) - 1/(-8.3)
1/v = -1/15.2 + 1/8.3
= -10/152 + 10/83
= (-830+1520)/12616
= 690/12616
v=12616/690=18.28cm