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
i wantere being if two different conventions are there in applications. Since I saw two different systems being followed.
To solve this problem, we can use the lens formula and magnification formula. Here's how you can calculate the image distance and height of the image:
1. Start by identifying the given values:
- Object distance (u) = -3cm (negative sign indicates that the object is placed in front of the lens)
- Focal length (f) = -6cm (negative sign indicates that the lens is diverging)
- Object height (h) = 1.5cm
2. Plug the values into the lens formula:
`1/f = 1/v - 1/u`
Substitute the given values:
`1/(-6) = 1/v - 1/(-3)`
Simplify the equation:
`-1/6 = 1/v + 1/3`
3. Multiply through by 6v to eliminate the denominators:
`-v = 6 + 2v`
Simplify further:
`3v = -6`
4. Solve for v (image distance):
`v = -6 / 3`
`v = -2cm`
Since the result is negative, it indicates that the image is formed on the same side as the object, making it a virtual image.
5. Calculate the height of the image (h'):
We can use the magnification formula: `h'/h = -v/u`
Substitute the given values:
`h'/1.5 = -(-2) / (-3)`
Simplify the equation:
`h'/1.5 = 2/3`
6. Solve for h' (height of the image):
`h' = (2/3) * 1.5`
`h' = 1cm`
Therefore, the image distance (v) is -2cm, and the height of the image (h') is 1cm.