• Home
  • Textbooks
  • Schaum’s Outline of College Physics
  • Optical Instruments

Schaum’s Outline of College Physics

Eugene Hecht

Chapter 39

Optical Instruments - all with Video Answers

Educators

DM

Chapter Questions

02:59

Problem 1

A nearsighted person named George cannot see distinctly objects beyond $80 \mathrm{~cm}$ from the eye. What is the power in diopters of the spectacle lenses that will enable him to see distant objects clearly?
The image, which must be right-side-up, must be on the same side of the lens as the distant object (hence, the image is virtual and $s_{i}=$
$-80 \mathrm{~cm}$ ), and nearer to the lens than the object (hence, diverging on negative lenses are indicated). Keep in mind that for virtual images formed by a concave lens $s_{0}>\left|s_{i}\right| .$ As the object is at a great distance, $s_{0}$ is very large and $1 / s_{0}$ is practically zero. Then
$$
\frac{1}{s_{n}}+\frac{1}{s_{i}}=\frac{1}{f} \quad \text { or } \quad 0-\frac{1}{80}=\frac{1}{f}
$$
$f=-80 \mathrm{~cm}$ (diverging?
and $\quad$ Power in diopters $=\frac{1}{f \text { in meters }}=\frac{1}{-0.80 \mathrm{~m}}=-1.3$ diopters

Vishal Gupta
Vishal Gupta
Numerade Educator
03:19

Problem 2

A farsighted person named Amy cannot see clearly objects closer to the eye than $75 \mathrm{~cm}$. Determine the power of the spectacle lenses which will enable her to read type at a distance of $25 \mathrm{~cm}$.

The image, which must be right-side-up, must be on the same side of the lens as the type (hence, the image is virtual and $s_{i}=75 \mathrm{~cm}$ ), and farther from the lens than the type (hence, converging on positive lenses are prescribed). Keep in mind that for virtual images formed by a convex lens $\left|s_{i}\right|>s_{0}$. We have
$$
\frac{1}{f}=\frac{1}{25}-\frac{1}{75} \quad \text { or } \quad f=+37.5 \mathrm{~cm}
$$
and
$$
\text { Power }=\frac{1}{0.375 \mathrm{~m}}=2.7 \text { diopters }
$$

Vishal Gupta
Vishal Gupta
Numerade Educator
04:06

Problem 3

A single thin projection lens of focal length $30 \mathrm{~cm}$ throws an image of a $2.0 \mathrm{~cm} \times 3.0 \mathrm{~cm}$ slide onto a screen $10 \mathrm{~m}$ from the lens. Compute the dimensions of the image.
The image is real and so $s_{i}>0$ :
and so 00
$$
\begin{array}{c}
\frac{1}{s_{o}}=\frac{1}{f}-\frac{1}{s_{i}}=\frac{1}{0.30}-\frac{1}{10}=3.23 \mathrm{~m}^{-1} \\
M_{T}=-\frac{s_{i}}{s_{o}}=-\frac{10 \mathrm{~m}}{(1 / 3.23) \mathrm{m}}=-32
\end{array}
$$
The magnification is negative because the image is inverted. The length and width of the slide are each magnified 32 times, so
Size of image $=(32 \times 2.0 \mathrm{~cm}) \times(32 \times 3.0 \mathrm{~cm})=64 \mathrm{~cm} \times 96 \mathrm{~cm}$

Vishal Gupta
Vishal Gupta
Numerade Educator
03:26

Problem 4

An old camera produces a clear image of a distant landscape when the thin lens is $8 \mathrm{~cm}$ from the film. What adjustment is required to get a good photograph of a map placed $72 \mathrm{~cm}$ from the lens?
When the camera is focused for distant objects (for parallel rays), the distance between lens and film is the focal length of the lens, namely, $8 \mathrm{~cm}$. For an object $72 \mathrm{~cm}$ distant:
$$
\frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{8}-\frac{1}{72} \quad \text { or } \quad s_{i}=9 \mathrm{~cm}
$$
The lens should be moved farther away from the film a distance of $(9-8) \mathrm{cm}=1 \mathrm{~cm}$

Vishal Gupta
Vishal Gupta
Numerade Educator
01:44

Problem 5

With a given illumination and film, the correct exposure for a camera lens set at $f / 12$ is $(1 / 5)$ s. What is the proper exposure time with the lens working at $f / 4$ ?

A setting of $f / 12$ means that the diameter of the opening, or stop, of the lens is $1 / 12$ of the focal length; $f / 4$ means that it is $1 / 4$ of the focal length.

The amount of light passing through the opening is proportional to its area, and therefore to the square of its diameter. The diameter of the stop at $f / 4$ is three times that at $f / 12$, so $3^{2}=9$ times as much light will pass through the lens at $f / 4$, and the correct exposure at $f / 4$ is
$$
(1 / 9)(\text { exposure time at } f / 12)=(1.45) \mathrm{s}
$$

Suzanne W.
Suzanne W.
Numerade Educator
02:54

Problem 6

An engraver who has normal eyesight uses a converging lens of focal length $8.0 \mathrm{~cm}$, which he holds very close to his eye. At what distance from the work should the lens be placed, and what is the magnification of the lens?
Method 1
When a converging lens is used as a magnifying glass, the object is between the lens and the focal point. The virtual erect, and
enlarged image forms at the distance of distinct vision, $25 \mathrm{~cm}$ from the eye. For a virtual image $s_{i}<0$. Thus,
Method 2
By the formula,
$$
M_{A}=\frac{d_{n}}{f}+1=\frac{25}{8.0}+1=4.1
$$
Note that in this simple case $M_{T}=M_{A}$.

Vishal Gupta
Vishal Gupta
Numerade Educator
07:28

Problem 7

Two positive lenses, having focal lengths of $+2.0 \mathrm{~cm}$ and $+5.0 \mathrm{~cm}$, are $14 \mathrm{~cm}$ apart as shown. An object $A B$ is placed $3.0$ $\mathrm{cm}$ in front of the $+2.0$ lens. Determine the position and magnification of the final image A"B" formed by this combination of lenses.
To locate image $A^{\prime} B^{\prime}$ formed by the $+2.0$ lens alone:
$$
\frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{2.0}-\frac{1}{3.0}=\frac{1}{6.0} \quad \text { or } \quad s_{i}=6.0 \mathrm{~cm}
$$
The image $A^{\prime} B^{\prime}$ is real, inverted, and $6.0 \mathrm{~cm}$ beyond the $+2.0$ lens.
To locate the final image $A^{\prime \prime} B^{\prime \prime}$ : The image $A^{\prime} B^{\prime}$ is $(14-6.0) \mathrm{cm}=$ $8.0 \mathrm{~cm}$ in front of the $+5.0$ lens and is taken as a real object for the
$+5.0$ lens.
$$
\frac{1}{s_{i}}=\frac{1}{5.0}-\frac{1}{8.0} \quad \text { or } \quad s_{i}=13.3 \mathrm{~cm}
$$
$A^{\prime \prime} B^{\prime \prime}$ is real, erect, and $13 \mathrm{~cm}$ from the $+5$ lens. Then,
$$
M_{T}=\frac{\overline{A^{\prime \prime} B^{\prime \prime}}}{\overline{A B}}=\frac{\overline{A^{\prime} B^{\prime}}}{\overline{A B}} \times \frac{\overline{A^{\prime \prime} B^{\prime \prime}}}{\overline{A^{\prime} B^{\prime}}}=\frac{6.0}{3.0} \times \frac{13.3}{8.0}=3.3
$$
Note that the magnification produced by a combination of lenses is the product of the individual magnifications.

Vishal Gupta
Vishal Gupta
Numerade Educator
07:07

Problem 8

In the compound microscope shown, the objective and eyepiece have focal lengths of $+0.80$ and $+2.5 \mathrm{~cm}$, respectively. The real intermediate image $A^{\prime} B^{\prime}$ formed by the objective is $16 \mathrm{~cm}$ from the objective. Determine the total magnification if the eye is held close to the eyepiece and views the virtual image $A^{\prime \prime} B^{\prime \prime}$ at a distance of $25 \mathrm{~cm}$.
Method 1
Let $s_{o} 0=$ Object distance from the objective
$s_{o} 0=$ Real-image distance from the objective
$$
\frac{1}{s_{o o}}=\frac{1}{f_{o}}-\frac{1}{s_{i O}}=\frac{1}{0.80}-\frac{1}{16}=\frac{19}{16} \mathrm{~cm}^{-1}
$$
and so the objective produces the linear magnification
$$
M_{T O}=-\frac{s_{i O}}{s_{o o}}=-(16 \mathrm{~cm})=\left(\frac{19}{16} \mathrm{~cm}^{-1}\right)=-19
$$
The intermediate image is inverted. The magnifying power of the eyepiece is
$$
M_{T E}=-\frac{s_{i E}}{s_{o E}}=-s_{i E}\left(\frac{1}{f_{E}}-\frac{1}{s_{i E}}\right)=-\frac{s_{i E}}{f_{E}}+1=-\frac{-25}{+2.5}+1=11
$$
The eyepiece does not flip the image: the intermediate image is inverted and the final image is inverted. Therefore, the magnifying power of the instrument is $-19 \times 11=-2.1 \times 10^{2}$.

Alternatively, under the conditions stated, the magnifying power of the eyepiece can be found as
$$
\frac{25}{f_{E}}+1=\frac{25}{2.5}+1=11
$$
Method 2
From Eq. (39.2) with $s_{i o}=16 \mathrm{~cm}$,
$$
\text { Magnification }=\left(\frac{d_{n}}{f_{E}}+1\right)\left(\frac{s_{i o}}{f_{o}}-1\right)=\left(\frac{25}{2.5}+1\right)\left(\frac{16}{0.8}-1\right)=2.1 \times 10^{2}
$$

Vishal Gupta
Vishal Gupta
Numerade Educator
02:20

Problem 9

The telephoto lens shown consists of a converging lens of focal length $+6.0 \mathrm{~cm}$ placed $4.0 \mathrm{~cm}$ in front of a diverging lens of focal length $-2.5 \mathrm{~cm} .(a)$ Locate the image of a very distant object. (b) Compare the size of the image formed by this lens combination with the size of the image that could be produced by the positive lens alone.
( $a$ ) If the negative lens were not employed, the intermediate image $A B$ would be formed at the focal point of the $+6.0$ lens, $6.0 \mathrm{~cm}$ distant from the $+6.0$ lens. The negative lens decreases the convergence of the rays refracted by the positive lens and causes them to focus at $A^{\prime} B^{\prime}$ instead of $A B$.
The image $A B$ (that would have been formed by the $+6.0$ lens alone) is $6.0-4.0=2.0 \mathrm{~cm}$ beyond the $-2.5$ lens and is taken as the (virtual) object for the $-2.5$ lens. Then $s_{o}=-2.0 \mathrm{~cm}$ (negative because $A B$ is virtual), and
$$
\frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{-2.5 \mathrm{~cm}}-\frac{1}{-2.0 \mathrm{~cm}}=\frac{1}{10 \mathrm{~cm}} \quad \text { or } \quad s_{i}=+10 \mathrm{~cm}
$$
The final image $A^{\prime} B^{\prime}$ is real and $10 \mathrm{~cm}$ beyond the negative lens.
(b) Magnification by negative lens $=\frac{\overline{A^{\prime} B^{\prime}}}{\overline{A B}}=-\frac{s_{i}}{s_{o}}=-\frac{10 \mathrm{~cm}}{-2.0 \mathrm{~cm}}=5.0$
so the diverging lens increases the magnification by a factor of $5.0 .$
Notice that the magnification produced by the convex lens is negative and so the net magnification of both lenses is negative:
the final image is inverted.

Suzanne W.
Suzanne W.
Numerade Educator
01:41

Problem 10

A microscope has two interchangeable objective lenses (3.0 mm and $7.0 \mathrm{~mm}$ ) and two interchangeable eyepieces $(3.0 \mathrm{~cm}$ and $5.0$
$\mathrm{cm}$ ). What magnifications can be obtained with the microscope if it is adjusted so that the image formed by the objective is $17 \mathrm{~cm}$ from that lens?
Because $s_{i o}=17 \mathrm{~cm}$ the magnification formula for a microscope, with $d_{n}=25 \mathrm{~cm}$, gives the following possibilities for $M_{A}$ :
For $f_{E}=3 \mathrm{~cm}, f_{o}=0.3 \mathrm{~cm}:$
For $f_{E}=3 \mathrm{~cm}, f_{o}=0.7 \mathrm{~cm}:$
For $f_{g}=5 \mathrm{~cm}, f_{\theta}=0.3 \mathrm{~cm}:$
For $f_{E}=5 \mathrm{~cm}, f_{o}=0.7 \mathrm{~cm}:$ $M_{A}=(9.33)(55,6)=518=5.2 \times 10^{2}$
$M_{A}=(9.33)(23.2)=216=2.2 \times 10^{2}$
$M_{A}=(5)(55.6)=278=2.8 \times 10^{2}$
$M_{A}=(5)(23.2)=116=1.2 \times 10^{2}$

Suzanne W.
Suzanne W.
Numerade Educator
02:11

Problem 11

Compute the magnifying power of a telescope, having objective and eyepiece lenses of focal lengths $+60$ and $+3.0 \mathrm{~cm}$, respectively, when it is focused for parallel rays.
The image is inverted.

Vishal Gupta
Vishal Gupta
Numerade Educator
02:43

Problem 12

Reflecting telescopes make use of a concave mirror, in place of the objective lens, to bring the distant object into focus. What is the magnifying power of a telescope that has a mirror with $250 \mathrm{~cm}$ radius and an eyepiece whose focal length is $5.0 \mathrm{~cm}$ ?
As it is for a refracting telescope (i.e., one with two lenses), $M_{A}=$ $-f_{O} / f_{E}$ again applies where, in this case, $f_{O}=-R / 2=125 \mathrm{~cm}$ and $f_{E}$ $=5.0 \mathrm{~cm} .$ Thus, $M_{A}=-25$.

Vishal Gupta
Vishal Gupta
Numerade Educator
08:06

Problem 13

As shown an object is placed $40 \mathrm{~cm}$ in front of a converging lens that has $f=+8.0 \mathrm{~cm} .$ A plane mirror is $30 \mathrm{~cm}$ beyond the lens. Find the positions of all images formed by this system.
For the lens
$$
\frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{8.0}-\frac{1}{40}=\frac{4}{40} \quad \text { or } \quad s_{i}=10 \mathrm{~cm}
$$
This is image $A^{\prime} B^{\prime}$ in the figure. It is real and inverted.
$A^{\prime} B^{\prime}$ acts as an object for the plane mirror, $20 \mathrm{~cm}$ away. A virtual image $C D$ is formed $20 \mathrm{~cm}$ behind the mirror.

Light reflected by the mirror appears to come from the image at $C D$. With $C D$ as object, the lens forms an image of it to the left of the lens. The distance $s_{i}$ from the lens to this latter image is given by
$$
\frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{8}-\frac{1}{50}=0.105 \quad \text { or } \quad s_{i}=9.5 \mathrm{~cm}
$$
The real images are therefore located $10 \mathrm{~cm}$ to the right of the lens and $9.5 \mathrm{~cm}$ to the left of the lens. (This latter image is upright.) A virtual inverted image is found $20 \mathrm{~cm}$ behind the mirror.

Vishal Gupta
Vishal Gupta
Numerade Educator
02:07

Problem 14

Two thin lenses having focal lengths of $+200 \mathrm{~cm}$ and $-400 \mathrm{~cm}$ are glued together so as to have a common axis. Determine the dioptric power of the combination. Discuss the physics of your answer. [Hint: Go back to Eq. (38.4), and then study the definition of dioptric power.]

Vishal Gupta
Vishal Gupta
Numerade Educator
04:41

Problem 15

A farsighted person who needs glasses can read without them when holding a text at $74 \mathrm{~cm}$ from the eye instead of the more usual $25 \mathrm{~cm}$. What eyeglass prescription does this person need? Discuss your answer.

Vishal Gupta
Vishal Gupta
Numerade Educator
02:17

Problem 16

A farsighted person who needs glasses can read without them
when holding a text at $74 \mathrm{~cm}$ from the eye instead of the more usual $25 \mathrm{~cm}$. If he or she puts on a pair of glasses having a dioptric power of $3.0 \mathrm{D}$, where will the new near point be? Discuss your answer. [Hint: Use Eq. (38.1) and remember that the image distance must be negative; here it's $-0.74 \mathrm{~m}$. The object distance is then the near point in meters.]

DM
Debra Mangion
Numerade Educator
03:44

Problem 17

A nearsighted person cannot see anything beyond $2.2 \mathrm{~m}$ very clearly. What eyeglass prescription does this person need? Discuss your answer. [Hint: Use Eq. (38.1) and remember that the image distance must be negative; here it's $-2.2 \mathrm{~m}$. The object distance is infinity.]

Vishal Gupta
Vishal Gupta
Numerade Educator
04:15

Problem 18

A farsighted person wears eyeglasses that provide a dioptric power of $+3.5 \mathrm{D}$. How far from his or her eyes must a book be held if it is to be read without using these glasses? Discuss your answer. [Hint:
Use Eq. (38.1) and remember that the image distance must be negative. Calculate the positive focal length of the lens. With glasses an object at $25 \mathrm{~cm}$ appears to be at $s i$, the near point, where it can be clearly seen by the eye.]

Vishal Gupta
Vishal Gupta
Numerade Educator
03:44

Problem 19

Redo the previous problem for someone wearing eyeglasses with a dioptric power of $+2.0$.

Vishal Gupta
Vishal Gupta
Numerade Educator
03:29

Problem 20

A farsighted woman cannot see objects clearly that are closer to her eye than $60.0 \mathrm{~cm}$. Determine the focal length and power of the spectacle lenses that will enable her to read a book at a distance of $25.0 \mathrm{~cm}$

Vishal Gupta
Vishal Gupta
Numerade Educator
02:49

Problem 21

A nearsighted man cannot see objects clearly that are beyond 50 $\mathrm{cm}$ from his eye. Determine the focal length and power of the glasses that will enable him to see distant objects clearly.

Vishal Gupta
Vishal Gupta
Numerade Educator
03:19

Problem 22

A projection lens is employed to produce $2.4 \mathrm{~m} \times 3.2 \mathrm{~m}$ pictures from $3.0 \mathrm{~cm} \times 4.0 \mathrm{~cm}$ slides on a screen that is $25 \mathrm{~cm}$ from the lens. Compute its focal length.

Vishal Gupta
Vishal Gupta
Numerade Educator
01:21

Problem 23

A camera gives a life-size picture of a flower when the thin lens is $20 \mathrm{~cm}$ from the film. What should be the distance between lens and film to photograph a flock of birds high overhead?

Suzanne W.
Suzanne W.
Numerade Educator
02:31

Problem 24

What is the maximum stop rating of a camera lens having a focal
length of $+10 \mathrm{~cm}$ and a diameter of $2.0 \mathrm{~cm} ?$ If the correct exposure at $f / 6$ is $(1 / 90)$ s, what exposure is needed when the diaphragm setting is changed to $f / 9 ?$

DM
Debra Mangion
Numerade Educator
02:11

Problem 25

What is the magnifying power of a lens of focal length $+2.0 \mathrm{~cm}$ when it used as a magnifying glass (or simple microscope)? The lens is held close to the eye, and the virtual image forms at the distance of distinct vision, $25 \mathrm{~cm}$ from the eye.

Vishal Gupta
Vishal Gupta
Numerade Educator
02:57

Problem 26

When the object distance from a converging lens is $5.0 \mathrm{~cm}$, a real image is formed $20 \mathrm{~cm}$ from the lens. What magnification is produced by this lens when it is used as a magnifying glass, the distance of most distinct vision being $25 \mathrm{~cm}$ ?

Vishal Gupta
Vishal Gupta
Numerade Educator
06:43

Problem 27

In a compound microscope, the focal lengths of the objective and eyepiece are $+0.50 \mathrm{~cm}$ and $+2.0 \mathrm{~cm}$, respectively. The instrument is focused on an object $0.52 \mathrm{~cm}$ from the objective lens. Compute the magnifying power of the microscope if the virtual image is viewed by the eye at a distance of $25 \mathrm{~cm}$.

Vishal Gupta
Vishal Gupta
Numerade Educator
03:28

Problem 28

A refracting astronomical telescope has a magnifying power of 150 when adjusted for minimum eyestrain. Its eyepiece has a focal length of $+1.20 \mathrm{~cm} .$ (a) Determine the focal length of the objective lens. (b) How far apart must the two lenses be so as to project a real image of a distant object on a screen $12.0 \mathrm{~cm}$ from the eyepiece?

Vishal Gupta
Vishal Gupta
Numerade Educator
02:48

Problem 29

The large telescope at Mt. Palomar has a concave objective mirror diameter of $5.0 \mathrm{~m}$ and radius of curvature $46 \mathrm{~m}$. What is the magnifying power of the instrument when it is used with an eyepiece of focal length $1.25 \mathrm{~cm}$ ?

Vishal Gupta
Vishal Gupta
Numerade Educator
01:08

Problem 30

An astronomical telescope with an objective lens of focal length $+80 \mathrm{~cm}$ is focused on the moon. By how much must the eyepiece be moved to focus the telescope on an object 40 meters distant?

Suzanne W.
Suzanne W.
Numerade Educator
04:47

Problem 31

A lens combination consists of two lenses with focal lengths of $+4.0 \mathrm{~cm}$ and $+8.0 \mathrm{~cm}$, which are spaced $16 \mathrm{~cm}$ apart. Locate and describe the image of an object placed $12 \mathrm{~cm}$ in front of the $+4.0$ cm lens.

DM
Debra Mangion
Numerade Educator
02:14

Problem 32

Two lenses, of focal lengths $+6.0 \mathrm{~cm}$ and $-10 \mathrm{~cm}$, are spaced $1.5$
$\mathrm{cm}$ apart. Locate and describe the image of an object $30 \mathrm{~cm}$ in front of the $+6.0$ -cm lens.

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
03:14

Problem 33

A telephoto lens consists of a positive lens of focal length $+3.5$ $\mathrm{cm}$ placed $2.0 \mathrm{~cm}$ in front of a negative lens of focal length $-1.8$ $\mathrm{cm} .$ ( $a$ ) 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.

Hunza Gilgit
Hunza Gilgit
Numerade Educator
01:27

Problem 34

An opera glass has an objective lens of focal length $+3.60 \mathrm{~cm}$ and a negative eyepiece of focal length $-1.20 \mathrm{~cm} .$ How far apart must the two lenses be for the viewer to see a distant object at $25.0 \mathrm{~cm}$ from the eye?

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
03:05

Problem 35

Repeat if the distance between the plane mirror and the lens is $8.0 \mathrm{~cm}$.

Abid Hussain
Abid Hussain
Numerade Educator
05:56

Problem 36

Solve if the plane mirror is replaced by a concave mirror with a $20 \mathrm{~cm}$ radius of curvature.

Abid Hussain
Abid Hussain
Numerade Educator