Determine the image distance and image height for a 5.0 CM tall object placed 45.0 cm from a concave mirror having the focal length of 15.0 CM
Answers
Given that, an object placed 45.0 cm from a concave mirror having the focal length of 15.0 cm and height of image is 5.0 cm.
We have to find the image distance (v) from the mirror and height of the image (hi).
Form above data we have, focal length (f) of concave mirror is -15 cm, object distance (u) is -45 cm and height of image (hi) is 5 cm.
Now, Using Mirror Formula:
1/f = 1/v + 1/u
Where f is focal length, v is image distance from the mirror and u is object distance from the mirror (concave).
Substitute the known values in the above formula to find the value of 'v' i.e. image distance from the mirror.
→ 1/(-15) = 1/v + 1/(-45)
→ -1/15 = 1/v - 1/45
→ -1/15 + 1/45 = 1/v
→ (-3 + 1)/45 = 1/v
→ -2/45 = 1/v
→ -45/2 = v
→ -22.5 = v
Therefore, the image distance from the concave mirror is -22.5 cm.
Now,
m = -v/u = hi/ho
Substitute the values,
→ -(-22.5)/(-45) = hi/5
→ -0.5 = hi/5
→ -2.5 = hi
Therefore, the height of the image is -2.5 cm.
Answer:
Given –
- Object Distance = u = - 45 cm
- Focal length = f = - 15 cm
- Object Size = h = 5 cm
- Concave mirror
To Find –
- Image Distance = v = ?
- Image Size = h' = ?
Solution –
Mirror Formula : 1/v + 1/u = 1/f
Substituting the values :
➸ 1/v + 1/-45 = 1/-15
➸ 1/v - 1/45 = -1/15
➸ 1/v = -1/15 + 1/45
➸ 1/v = (-45+15)/675
➸ 1/v = -30/675
➸ 1/v = -6/135
➸ -6v = 135
➸ v = -135/6
➸ v = -22.5 cm
Image Distance = -22.5 cm
Now, Magnification = -v/u = h'/h
➸ Magnification = -(-22.5)/-45 = h'/5
➸ -225/450 = h'/5
➸ -225 * 5 = 450 * h'
➸ (-225 * 5)/450 = h'
➸ -225/90 = h'
➸ h' = -2.5 cm
Image Size = -2.5 cm
Hence,
➜ The Image will be Diminished. [Size of the Image]
➜ The Image will be real and inverted. [Nature of Image]
➜ The Image formed will be between Focus and Centre of Curvature. [Position of the Image]
