Chapter 39, Problem 33SP

A telephoto lens consists of a positive lens of focal length +3.5 cm placed 2.0 cm in front of a negative lens of focal length –1.8 cm. (a) Locate the image of a very distant object. (b) Determine the focal length of the single lens that would form as large an image of a distant object as is formed by this lens combination.

To determine

The distance of the image of an object, which is very far away from the telephoto lens, if it has a positive focal length of $+3.5\text{ cm}$ and is placed $2.0\text{ cm}$ in front of a negative lens of focal length $-1.8\text{ cm}$.

Answer to Problem 33SP

Solution:

A real image at a distance of $9.0\text{ cm}$ is formed behind the negative lens.

Explanation of Solution

Given data:

The focal length of the negative lens is $-1.8\text{ cm}$.

The focal length of the positive lens is $3.5\text{ cm}$.

The positive lens is placed $2.0\text{ cm}$ in front of the negative lens.

Formula used:

The expression for the thin lens formula is written as

$\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f}$

Here, $s_i$ is the distance of the image, $s_o$ is the distance of the object from the lens, and $f$ is the focal length of the lens.

Sign convention:

$s_o$ is taken to be positive for the real object when the object is at the left side of the lens, and negative for virtual object.

$s_i$ is positive when the image is at the right side of the lens and negative when it is virtual and formed left to the lens.

$f$ is positive for converging lens and negative for diverging lens.

Explanation:

Draw the diagram according to the problem.

Here, $A$ is the positive lens, $B$ is the negative lens, $f_A$ is the focal length of the convex lens, $f_B$ is the focal length of the concave lens, and $D$ is the distance between both the lenses.

Understand that for the given diagram, the image of the object at very distant formed at the focus of the positive lens, which works as an object for the negative lens. So, the distance of the object from the negative lens will be the difference between its focal length and the distance between both the lenses.

The expression for the distance of the object from a negative lens is

$s_o=f_A-D$

Substitute $3.5\text{ cm}$ for $f_A$ and $2.0\text{ cm}$ for $D$

$\begin{array}{c}{s}_o=3.5\text{ cm}-2.0\text{ cm}\\ =1.5\text{ cm}\end{array}$

Also, the object will be at right side for negative lens. So, its distance will be negative according to the sign convention. Therefore,

$s_o=-1.5\text{ cm}$

Write the expression for the lens formula for a thin lens:

$\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f_B}$

Substitute $-1.5\text{ cm}$ for $s_o$ and $-1.8\text{ cm}$ for $f_B$

$\begin{array}{c}\frac{1}{s_i}+\frac{1}{\left(-1.5\text{ cm}\right)}=\frac{1}{-1.8\text{ cm}}\\ \frac{1}{s_i}=\frac{1}{-1.8\text{ cm}}+\frac{1}{1.5\text{ cm}}\\ s_i=9.0\text{ cm}\end{array}$

The positive sign of the distance of the image shows that a real image is formed $9.0\text{ cm}$ behind the negative lens.

Conclusion:

Hence, the image formed of a very distant object from the telephoto is real and $9.0\text{ cm}$ behind the negative lens.

To determine

The focal length of a single lens that will form as large image of distant object as formed by the combination of a positive lens of focal length $+3.5\text{ cm}$ placed $2.0\text{ cm}$ in front of a negative lens of focal length $-1.8\text{ cm}$.

Answer to Problem 33SP

Solution:

$+21\text{ cm}$

Explanation of Solution

Given data:

The focal length of the positive lens is $3.5\text{ cm}$.

The focal length of the negative lens is $-1.8\text{ cm}$.

The positive lens is placed $2.0\text{ cm}$ in front of the negative lens.

Formula used:

The expression for the resultant focal length of a combination of lenses is written as

$\frac{1}{F}=\frac{1}{f_A}+\frac{1}{f_B}-\frac{D}{f_A{f}_B}$

Explanation:

Recall the expression for the resultant focal length of a combination of lenses:

$\frac{1}{F}=\frac{1}{f_A}+\frac{1}{f_B}-\frac{D}{f_A{f}_B}$

Substitute $3.5\text{ cm}$ for $f_A$, $-1.8\text{ cm}$ for $f_B$, and $2.0\text{ cm}$ for $D$

$\begin{array}{c}\frac{1}{F}=\frac{1}{3.5\text{ cm}}+\frac{1}{\left(-1.8\text{ cm}\right)}-\frac{2.0\text{ cm}}{\left(3.5\text{ cm}\right)\left(-1.8\text{ cm}\right)}\\ \frac{1}{F}=0.0476{\text{ cm}}^{-1}\\ F=+21\text{ cm}\end{array}$

Conclusion:

The focal length of the single lens will be $+21\text{ cm}$.