An object is placed 3cm in front of a diverging lens of focal length of 6cm. the heihgt of the object is 1.5 cm. calculate the image distance and height of the image.
-1/6-1/3=-1/2
image distance=-2
hi/ho=-di/do
hi/1.5=-(-2)/3
hi=(2/3)x(3/2)
hi=1
To calculate the image distance and height of the image formed by a diverging lens, we can use the lens formula and the magnification formula.
1. Lens Formula:
The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens. It is given by:
1/f = 1/v - 1/u
where:
f = focal length of the lens
v = image distance
u = object distance (distance of the object from the lens)
2. Magnification Formula:
The magnification formula relates the height of the image (h') and the height of the object (h). It is given by:
magnification (m) = h'/h = -v/u
where:
h' = height of the image
h = height of the object
v = image distance
u = object distance
Given:
focal length (f) = -6 cm (Since it is a diverging lens, focal length is negative)
object distance (u) = -3 cm (Since the object is in front of the lens, the distance is negative)
height of the object (h) = 1.5 cm
Now, let's calculate the image distance (v):
1/f = 1/v - 1/u
Rearranging the formula and substituting the values:
1/v = 1/f + 1/u
1/v = 1/-6 + 1/-3
1/v = -1/6 - 1/3
1/v = -1/6 - 2/6
1/v = -3/6
v = -6/3
v = -2 cm
Therefore, the image distance (v) is -2 cm (since it is negative).
Now, let's calculate the height of the image (h'):
magnification (m) = h'/h = -v/u
Substituting the values:
m = h'/h
-2/3 = h'/1.5
Cross-multiplying and solving for h':
-2/3 * 1.5 = h'
h' = -1
Therefore, the height of the image (h') is -1 cm.
Therefore, the image distance is -2 cm and the height of the image is -1 cm.