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A $0.75\, {\rm m}$ footstool is placed in front of a lens. On a screen $4.00\,{\rm m}$ away from the stool is displayed the stool's image. The image of the stool is $3.00\,{\rm m}$ tall on the screen.

(a) Is the lens used a positive or negative lens? Characterize the image formed on the screen.

(b) Determine $d_i\,\ d_o$ and $f$ for this lens.

(c) Construct a ray-tracing diagram for the situation.

(i) Converging lens have positive focal length. If $d_o>f$ then the image distance $d_i$ is positive (that is, image is formed on the other side of lens, inverted and real). If $d_o<f$ then $d_i<0$ (that is, object and image is located in the same side, virtual and larger)

(ii) Diverging lens have negative focal length and object and image are in the same side so $d_i<0$.

This lens has properties of the converging lens due to the larger image and image formed on the other side of the lens.

(b) The object-image relationship for lens are similar to the spherical mirrors.

\[\frac{1}{d_i}+\frac{1}{d_o}=\frac{1}{f}\ \ \ {\rm and\ \ }\ M=-\frac{d_i}{d_o}=\frac{h_i}{h_o}\]

In this problem $d_i=L-d_o$ where $L$ is the distance between image and object so

\[\left|-\frac{L-d_o}{d_o}\right|=\frac{h_i}{h_o}=\frac{0.75}{3}=4\ \Rightarrow L-d_o=4d_o\]

\[4-d_o=4d_o\to d_o=0.8{\rm m\ \ and\ \ }{{\rm d}}_{{\rm i}}{\rm =L-}{{\rm d}}_{{\rm o}}==3.2\ {\rm m}\]

\[\frac{1}{d_i}+\frac{1}{d_o}=\frac{1}{f}\to f=\frac{d_od_i}{d_o+d_i}=\frac{0.8\times 3.2}{3.2+0.8}=0.64\ {\rm m}\]

Note: the object is placed at $f<d_o<2f$, just what is needed for real and inverted image on screen.

(c)

An object is placed $10\, {\rm cm}$ to the left of a diverging lens ($f=-5\,{\rm cm}$). A concave mirror ($R=20\,{\rm cm}$) is placed $20\,{\rm cm}$ to the right of the lens. Calculate the location of the final image with respect to the lens.

Object $O_1$ is $15\, {\rm cm}$ to the left of a converging lens of $10\, {\rm cm}$ focal length. A second lens is positioned $10\, {\rm cm}$ to the right of the first lens and is observed to form a final image at the position of the original object $O_1$.

(b) What is the focal length of the second lens?

(b) What is the overall magnification of this system?

(c) Is the final image real or virtual, upright or inverted? Justify your answer.

A $10\,{\rm cm}$ tall candle is placed $40\,{\rm cm}$ in front of a concave mirror of radius $R=60\,{\rm cm}$.

(a) Determine the position and size of the image of the candle formed by this mirror.

(b) Characterize this image and draw a ray diagram on the above figure.

(c) A second mirror (convex) whose radius $R=60\,{\rm cm}$ is placed $10\,{\rm cm}$ to the left of the candle. The convex surface is the surface facing the candle. Determine the position and size of the image of the candle formed by the convex mirror.

(d) Determine the position and size of the image of the candle formed by the convex mirror coming from the rays reflected from the concave mirror.

A spherical mirror is used to form an image, five times as tall as an object, on a screen positioned $5.0\,{\rm m}$ from the mirror.

(a) Describe the mirror (convex or concave, radius of curvature and focal length).

(b) Where the mirror should be placed relative to the object?

An optometrist prescribes contact lens that have a focal length of $55\,{\rm cm}$.

(a) Are the lenses converging or diverging? Justify your answer.

(b) Is the person wearing the lenses nearsighted or farsighted?

(c) Where is the unaided near point of the person located, if the lenses are designed so that objects no closer than $35\,{\rm cm}$ can be seen clearly?

A farsighted person has a near point that is $48\,{\rm cm}$ from her eyes. She wears eyeglasses that are designed to enable her to read a newspaper held at a distance of $26\,{\rm cm}$ from her eyes. Find the focal length of the eyeglasses if they are worn $2\,{\rm cm}$ from the eyes.

A farsighted person can see distant objects clearly and farsightedness is corrected with a converging (positive) lens.

Note: For a farsighted person, a sharp image of an object located between the near point and the eye forms behind the retina. With a converging lens, this image is moved onto the retina. Recall that the image of a first lens acts as the object for a second lens. In the lens-eye system, the eye is the second lens. The goal is the person can focus on an object located at near point (normal vision). A converging lens forms an upright, virtual image at the person’s actual near point($NP=26 \,{\rm cm}$). This image acts as an object for the eye.

Since after wearing the glasses, she able to read normally so she has gained the normal vision. In general, normal vision will allow a person to focus on an object $26\,{\rm cm}$ away. The $48\, {\rm cm}$ near point of this person means she can see objects at $48\,{\rm cm}$. using maximum accommodation, we must use a lens that creates a virtual image at position $d_i=-46\,{\rm cm}$ (in lenses if image and object are in the same side then $d_i$ is virtual and negative)from an object at $d_o=24\,{\rm cm}$.

\[\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\ \Rightarrow f=\frac{d_id_o}{d_i+d_o}=\frac{-46\times 24}{-46+24}=+50.2\,{\rm cm}\]

The positive sign indicates that the lens is converging, as expected.

Josh is farsighted and has a near point located $119\,{\rm cm}$ from his eyes. Anne is also farsighted, but her near point is $72\,{\rm cm}$ from her eyes. Both have glasses that correct their visions to a normal near point ($35\,{\rm cm}$ from the eyes), and both wear glasses $3.0\,{\rm cm}$ from the eyes. Relative to the eyes, what is the closet object that can be seen by Josh when he wears Anne's glasses?

Two identical diverging lenses are separated by $18\,{\rm cm}$. The focal length of each lens is $-8.0\,{\rm cm}$. A $6{\rm cm}$ tall object is located $4.0\,{\rm cm}$ to the left of the lens that is on the left. Clearly label all images.

(a) Construct the ray diagram.

(b) Determine the final image distance relative to the lens on the right.

(c) How tall is the final image?

A candle is placed $12\,{\rm cm}$ in front of a convex mirror. When the convex mirror is replaced with a plane mirror, the image moves $8.5\,{\rm cm}$ farther away from the mirror. Find the focal length of the convex mirror.

Plane mirror has exact reflection i.e. $d_i=-d_o$. Let $d_i$ and $d_{i_2}$as the image distances of convex and plane mirrors, respectively. So as shown in the figure, we have

\[d_{i_2}-d_i=8.5\,{\rm cm}\to {{\rm d}}_{{\rm i}}=12-8.5=3.5\,{\rm cm}\]

Use the mirror equation to find the focal length of the convex mirror. Note that the image distance in the convex mirror is negative since image is virtual!

\[\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}=\frac{1}{-3.5}+\frac{1}{12}\ \Longrightarrow f=-4.94\,{\rm cm}\]

Two identical converging lenses with $f=30\,{\rm cm}$ are $70\,{\rm cm}$ apart. A $12\,{\rm cm}$ tall object is placed $100\,{\rm cm}$ in front of the first lens. What is the final height of the object (including sign)?

An object is located $14\,{\rm cm}$ in front of a concave mirror, with the image $7\,{\rm cm}$ in front of the mirror. A second object, one third as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height (in magnitude only) as the first image, but the orientation is flipped. How far in front of the mirror, is the second object located?

An object is standing to the left of a diverging lens ($f=-8\, {\rm cm}$). A converging lens ($f=6.3\,{\rm cm}$) is $21\,{\rm cm}$ to the right of the diverging lens. The image created by the diverging lens is a distance of $5\,{\rm cm}$ from the diverging lens.

(a) How far is the final image from the object?

(b) What is the height of the final image if the object is $2.0\,{\rm cm}$ tall?

(c) Is the final image virtual or real?

(d) Is the final image inverted or upright?

An object sits to the left of a lens at a distance of $74.2\, {\rm cm}$. An inverted image is formed on the right of the lens that is $2.88$ times the size of the object.

(a) Where is the image formed with respect to the lens and is it real or virtual?

(b) What is the focal length of the lens and is this a diverging or converging lens?

Given data: $d_o=74.2\, {\rm cm\ ,\ }h_i=-2.88h_o$. (The negative inserted since the image is inverted). From the magnification equations, we have:

\[-\frac{d_i}{d_o}=\frac{h_i}{h_o}\Rightarrow d_i=-d_o\frac{h_i}{h_o}=-74.2\ \left(\frac{-2.88h_i}{h_i}\right)=213.696\ {\rm cm\ \sim +}214\, {\rm cm}\]

In lenses if the image and object are in the same side then $d_i<0\ $and the image is virtual otherwise is real. Since $d_i>0$ so the image is __real__.

(b) Use the mirror equation to find the focal length

\[\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\Rightarrow \frac{1}{f}=\frac{1}{214}+\frac{1}{74.2}=0.0181\Rightarrow f=55.09\, {\rm m}\]

Since $f>0$ the lens is __converging__.

For an object of height $9.00\,{\rm cm}$ located at a distance $1.1\,{\rm m}$ from a mirror, find the height of the image and its distance to the mirror for each of following three cases. In each case also state whether the image is in front or behind the mirror and whether the image is upright or inverted.

(a) Convex mirror with radius of curvature of $R=1.6\,{\rm m}$

(b) Concave mirror with radius of curvature of $R=1.6\,{\rm m}$

(c) Plane mirror.

A single concave spherical mirror is used to create an image of a $5\,{\rm cm}$ tall source that is located at position$x=0\,{\rm cm}$, which is $20\,{\rm cm}$ to the left point C, the center of the mirror, as shown in the figure below. The magnitude of the radius of curvature for the mirror is $10{\rm cm}$.

(a) What is the source distance $s$ from the mirror?

(b) What is the magnitude and sign of the mirror's focal length?

(c) Using three principal rays, show the resulting image?

(d) Calculate the position $x_i$ where the image is formed?

(d) What is the height $h_i$ of the image?

(e) Is the image upright or inverted? Is it real or virtual?

The image formed by a convex mirror ($R=70\,{\rm cm}$) is located $6\,{\rm cm}$ from the mirror.

(a) What is the image distance?

(b)What is the object distance?

(c) If the height of the image if $4.84\, {\rm cm}$, what is the height of the object?

(d) Characterize the image.

A concave make-up mirror produces an erect image that is $2.5$ times the size of the object when the object is placed $12\,{\rm cm}$ in front of the mirror.

(a) Determine the focal length of the mirror.

(b) Determine the image distance for this setup.

(c) Use an arrow for the object and construct a ray tracing for the situation above.

\[\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\to f=\frac{d_id_0}{d_i+d_o}=\frac{-2.5d_o\times d_o}{-2.5d_o+d_o}=\frac{-2.5\times 12}{-1.5}=20\,{\rm cm}\]

(b) Using the magnification equation, we obtain

\[d_i=-Md_o=\left(-2.5\right)\left(12\right)=-30\,{\rm cm}\]

Since $d_i<0$ so the image is virtual.

(c) Apply the principles of the drawing the ray diagrams.

Category : Optics

MOST USEFUL FORMULA IN OPTICS:

Lateral magnification in any reflecting or refracting situation:

\[m=\frac{image\ height}{object\ height}=\frac {h_i}{h_o}\]

$m>0 \Longrightarrow$ image is erect.

$m<0 \Longrightarrow$ image is inverted.

Object-Image relationship for mirrors and thin-lenses:

\[\frac 1 s+\frac {1}{s^{'}}=\frac 1 f\]

where $s$ and $s^{'}$ are the object and image distances, respectively.

Focal length and radius of curvature are related as

\[f=\frac R 2\]

Number Of Questions : 18

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