Chapter 23, Problem 94GP

To determine

The focal length of the diverging lens.

Answer to Problem 94GP

Solution:

19∙01 cm.

Explanation of Solution

Object is placed behind the diverging (concave)lens at 25 cm. A converging (convex)lens whose focal length is 12 cm, is placed at 30 cm to the right from the diverging (concave)lens. Final image formed by two lens systems is at 17 cm to the right of converging (convex)lens and the image is real and inverted.

Given:

Object distance from concave lens = 25 cm

Focal length of convex lens = 12 cm

Distance between both lenses is = 30 cm

Distance of final image from convex (converging)lens = 17 cm

Formula used:

Lens formula for focal length

$1/f=1/v-1/u$

Calculation:

For converging lens

v=17 cm, f=12 cm, u=?

$\begin{array}{l}1/f=1/v-1/u\hfill \\ 1/u=1/v–1/f\hfill \\ 1/u=\left(1/17\right)–\left(1/12\right)=\left(12-17\right)/\left(12\times 17\right)=-5/204\hfill \\ u=-204/5=-40\cdot 8\text{ cm }\left(\text{- sign means behind the lens}\right)\hfill \end{array}$

so now for diverging lens v = (40∙8-30)cm = 10∙8 cm behind the diverging lens.

Now, v = -10∙8 cm, u = -25 cm,

$\begin{array}{l}\begin{array}{l}1/f=1/v–1/u=\left(-1/10\cdot 8\right)–\left(-1/25\right)\\ =\left(1/25\right)-\left(1/10\cdot 8\right)\\ =\left(10\cdot 8–25\right)/\left(25\times 10\cdot 8\right)\end{array}\hfill \\ 1/f=\left(-14\cdot 2/270\right)\hfill \\ f=-270/14\cdot 2=-19\cdot \le 01\mathrm{cm}\hfill \end{array}$