Question

to practice problem-solving strategy 34.2 image formation by thin lenses. a 2 cm high object is located 30 cm from a diverging lens, whose focal length is 13 cm . what is the height of the image produced by the lens?

Answer 1

The height of the image produced by the diverging lens is approximately
`\(1.46 \, \text{cm}\)`

To solve this problem, you can use the thin lens formula, which relates the object distance
`(\(u\))`, the image distance
`(\(v\))`, and the focal length
`(\(f\))`of a lens:

`\[ (1)/(f) = (1)/(v) - (1)/(u) \]`

In this case, the object distance
`(\(u\))` is given as -30 cm (since the object is on the same side as the diverging lens), and the focal length
`(\(f\))` is given as 13 cm (negative for a diverging lens).

1. Identify Known Values:

-
`\(u = -30 \, \text{cm}\)` (object distance)

-
`\(f = -13 \, \text{cm}\)` (focal length of the diverging lens)

2. Apply the Thin Lens Formula:

`\[ (1)/(f) = (1)/(v) - (1)/(u) \]`

3. Solve for \(v\):

`\[ (1)/((-13)) = (1)/(v) - (1)/((-30)) \]`

4. Calculate
`\(v\)`:

`\[ (1)/(v) = (1)/((-13)) + (1)/((-30)) \]`

5. Solve for
`\(v\)`:

`\[ v = (1)/(\left((1)/((-13)) + (1)/((-30))\right)) \]`

6. Calculate
`\(v\)`:

Plug in the values and calculate.

`\[ v \approx -21.82 \, \text{cm} \]`

Now that you have the image distance
`(\(v\))`, you can use the magnification formula to find the height of the image
`(\(h'\))` in terms of the object height
`(\(h\))`:

`\[ (h')/(h) = -(v)/(u) \]`

7. Calculate
`\(h'\)`:

`\[ h' = \left(-(v)/(u)\right) \cdot h \]`

`\[ h' = \left(-((-21.82))/((-30))\right) \cdot 2 \, \text{cm} \]`

8. Calculate
`\(h'\)`:

`\[ h' \approx 1.46 \, \text{cm} \]`

So, the height of the image produced by the diverging lens is approximately
`\(1.46 \, \text{cm}\)`.