Chapter 23, Problem 64P

Two lenses, one converging with focal length 20.0 cm and one diverging with focal length -10.0 cm. are placed 25.0 cm apart. An object is placed 60.0 cm in front of the converging lens. Determine (a) the position and (b) the magnification of the final image formed, (c) Sketch a ray diagram for this system.

To determine

(a)

The position of the image.

Answer to Problem 64P

Solution:

The image forms $10cm$ in the right of second lens.

Explanation of Solution

Given:

Converging lens has focal length $20.0cm$ and the diverging one has focal length $-10.0cm$. They are placed $25cm$ apart. Object is placed $60cm$ before the first lens.

Formula used:

Lens equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$f,u$ and $v$ are lens to focal, lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.

Magnification, $m=\left|\frac{v}{u}\right|$

Calculation:

$u_1=-60$ cm

$f_1=20$ cm

Image distance formed by first lens $v_1$

Using lens equation,

$\frac{1}{v_1}+\frac{1}{60}=\frac{1}{20}$

This gives, $v_1=30$ cm

Now, the distance between two lenses is 25 cm.

The image formed by first lens acts as a virtual object for second lens.

Thus, object distance for second lens is $u_2=30-25=5$ cm

Considering final image distance from second lens as $v_2$ we get,

$\frac{1}{v_2}-\frac{1}{5}=\frac{1}{f_2}$

Focal length of the diverging lens $f_2=-10$ cm

To find the position substitute the values to lens equation, we get,

$\frac{1}{v_2}-\frac{1}{5}=-\frac{1}{10}$

$\begin{array}{l}\frac{1}{v_2}=\frac{1}{5}-\frac{1}{10}\\ \frac{1}{v_2}=\frac{1}{10}\end{array}$

$v_2=10cm$

To determine

(b)

Magnification of the image formed by two lenses, one converging with focal length $20.0cm$ and one diverging with focal length $-10.0cm$ placed $25cm$ apart with object placed $60cm$ in front of converging lens.

Answer to Problem 64P

Solution:

The magnification is $1$.

Explanation of Solution

Given:

Converging lens has focal length $20.0cm$ and the diverging one has focal length $-10.0cm$. They are placed $25cm$ apart. Object is placed $60cm$ before the first lens.

Formula used:

Lens equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$f,u$ and $v$ are lens to focal, lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.

Magnification, $m=\left|\frac{v}{u}\right|$

Calculation:

${\text{u}}_{\text{1}}\text{=-60}$ cm

$f_1=20$ cm

Image distance formed by first lens $v_1$

Using lens equation,

$\frac{1}{v_1}+\frac{1}{60}=\frac{1}{20}$

This gives, $v_1=30$ cm

Now, the distance between two lenses is 25 cm.

The image formed by first lens acts as a virtual object for second lens.

Thus, object distance for second lens is $u_2=30-25=5$ cm

Considering final image distance from second lens as $v_2$ we get,

$\frac{1}{v_2}-\frac{1}{5}=\frac{1}{f_2}$

Focal length of the diverging lens $f_2=-10$ cm

Thus, the total magnification is $m=\left|\frac{v_2{v}_1}{u_2{u}_1}\right|$=1.

To determine

(c)

The ray diagram of the system.

Answer to Problem 64P

Solution:

Explanation of Solution

Given:

Converging lens has focal length $20.0cm$ and the diverging one has focal length $-10.0cm$. They are placed $25cm$ apart. Object is placed $60cm$ before the first lens.

Formula used:

Lens equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$f,u$ and $v$ are lens to focal, lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.

Magnification, $m=\left|\frac{v}{u}\right|$

Calculation:

$u_1=-60$ cm

$f_1=20$ cm

Image distance formed by first lens $v_1$

Using lens equation,

$\frac{1}{v_1}+\frac{1}{60}=\frac{1}{20}$

This gives, $v_1=30$ cm

Now, the distance between two lenses is 25 cm.

The image formed by first lens acts as a virtual object for second lens.

Thus, object distance for second lens is $u_2=30-25=5$ cm

Considering final image distance from second lens as $v_2$ we get,

$\frac{1}{v_2}-\frac{1}{5}=\frac{1}{f_2}$

Focal length of the diverging lens $f_2=-10$ cm

Ray diagram of the system is as below-