Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)
Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)
9th Edition
ISBN: 9781305116429
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 36, Problem 36.50P

In Figure P35.30, a thin converging lens of focal length 14.0 cm forms an image of the square abed, which is he = hb = 10.0 cm high and lies between distances of pd = 20.0 cm and pa = 30.0 cm from the lens. Let a′, b′, c′. and d′ represent the respective corners of the image. Let qa represent the image distance for points a′ and b′, qd represent the image distance for points c′ and d′, h b , represent the distance from point b′ to the axis, and h c represent the height of c′. (a) Find qa, qd, h b , and h c . (b) Make a sketch of the image. (c) The area of the object is 100 cm2. By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a′ and d′, for which the object distance is p. Let h′ represent the distance from the axis to the point at the edge of the image between b′ and c′ at image distance q. Demonstrate that

| h | = 10.0 q ( 1 14.0 1 q )

where h′ and q are in centimeters. (d) Explain why the geometric area of the image is given by

q a q d | h | d q

(e) Carry out the integration to find the area of the image.

Figure P35.30

Chapter 36, Problem 36.50P, In Figure P35.30, a thin converging lens of focal length 14.0 cm forms an image of the square abed,

(a)

Expert Solution
Check Mark
To determine
The position of the given constants. qa , qd , hb and hc .

Answer to Problem 36.50P

The values of qa , qd hb and hc is 26.3cm , 46.7cm , 8.75cm and 23.3cm respectively

Explanation of Solution

Given info: The focal length of the lens is 14.0cm . The object is square abcd of height 10.0cm and the positioned between 20cm and 30.0cm .

Formula to calculate the image of any object in a thin lens is,

1f=1q+1p (1)

Here,

f is the focal length of the lens.

p is the object position.

q is the image position.

Substitute qa for q , pa for p in the equation (1),

1f=1pa+1qa (2)

Here,

pa is the object distance corresponding to ab .

qa is the image distance corresponding to ab .

Substitute 30.0cm for pa and 14.0cm for f in equation (2)

1f=1pa+1qa114.0cm=130.0cm+1qaqa=26.3cm

Substitute pd for p and qd for q in the equation (1),

1f=1qd+1pd (3)

Here,

qd is the image distance corresponding to the cd .

pd is the object position corresponding to cd .

Substitute, 20.0cm for pd and 14.0cm for f in equation (3),

1f=1qd+1pd114.0cm=1qd+120.0cmpd=46.7cm

Formula to calculate the height of the corresponding image point ab

Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics), Chapter 36, Problem 36.50P , additional homework tip  1 hb=hb(qapa) (4)

Here,

hb is to the height of image corresponding to ab

hb is the height of the object corresponding to ab .

Substitute 10.0cm for hb , 26.3cm for qa and 30.0cm in equation (4).

hb=hb(qapa)hb=10.0cm(26.3cm30.0cm)hb=8.75cm

 For calculating the image height corresponding to cd ,

hc=hc(qdpd) (5)

Substitute 10.0cm for hd , 46.7cm for qa and 20.0cm in equation (5).

hc=hc(qdpd)hc=10.0cm(46.7cm20.0cm)hc=23.3cm

Conclusion:

Therefore, the values of qa , qd hb and hc is 26.3cm , 46.7cm , 8.75cm and 23.3cm respectively.

(b)

Expert Solution
Check Mark
To determine

To draw: The sketch of the ray diagram.

Answer to Problem 36.50P

 The sketch of the ray diagram is,

Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics), Chapter 36, Problem 36.50P , additional homework tip  2

Explanation of Solution

The image of the square abcd will be a trapezoid of abcd .

Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics), Chapter 36, Problem 36.50P , additional homework tip  3

Figure (1)

(c)

Expert Solution
Check Mark
To determine

To show: The relation, |h|=10q(114.0cm1q) .

Answer to Problem 36.50P

The relation between height of the image and the image distance is |h|=10q(114.0cm1q) .

Explanation of Solution

the focal length of the given lens is 14.0cm . The formula to calculate the magnification is

hh=qp (6)

Here,

h is the image height.

h is the object height.

q is the image distance.

p is the object distance.

From the lens,

1f=1p+1q (7)

Here,

f is the focal length of the lens.

Substitute 14.0cm for f in equation (7).

1f=1p+1q114.0cm=1p+1q1p=114.0cm1q (8)

Substitute 114.0cm1q for p and 10.0cm for h in equation (6),

hh=qphh=q(1p)h=10(114.0cm1q)q|h|=10q(114.0cm1q) (9)

Conclusion:

Therefore the relation between height of the image and the image distance is |h|=10q(114.0cm1q) .

(d)

Expert Solution
Check Mark
To determine

To write: The explanation that the geometric area of image is qaqd|h|dq

Answer to Problem 36.50P

The geometric area of a image is explained by the integral qaqd|h|dq .

Explanation of Solution

Given info: The geometric area of a image is,

Area=qaqd|h|dq . (10)

Here,

h is the height of the image.

dq is the smaller region of image distance.

From equation (10) the integral sums up the small areas of region covered by the image itself. The height of the small regions is h and the small portion of image distance is dq .

Therefore area of that small region is

dA=hdq

Therefore the integration from qa   to qd of dA will give the geometric area of the image.

Conclusion:

Therefore, the geometric area of the image is given by integral qaqd|h|dq .

(e)

Expert Solution
Check Mark
To determine
The geometric area of the image.

Answer to Problem 36.50P

The geometric area is 328cm2 .

Explanation of Solution

Given info: The geometric area of the image is given by the integral,

qaqd|h|dq . (11)

From equation (9) substitute 10q(114.0cm1q) for h and 26.3cm for qa , 46.7cm for qd in equation (11)

qaqd|h|dq26.3cm46.7cm10q(114.0cm1q)dq

Integrate the above equation with respect to dq to get the area ,

Area=26.3cm46.7cm10q(114.0cm1q)dq=10.0cm[q228.0cmq]26.3cm46.7cm=10.0cm[[(46.7cm)226.3cm2]28.0cm46.7cm+26.3cm]=328cm2

Conclusion:

Therefore, the geometric area of the image is 328cm2 .

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Chapter 36 Solutions

Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)

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At a...Ch. 36 - Prob. 36.17PCh. 36 - A certain Christmas tree ornament is a silver...Ch. 36 - (a) A concave spherical mirror forms an inverted...Ch. 36 - (a) A concave spherical mirror forms ail inverted...Ch. 36 - An object 10.0 cm tall is placed at the zero mark...Ch. 36 - A concave spherical mirror has a radius of...Ch. 36 - A dedicated sports car enthusiast polishes the...Ch. 36 - A convex spherical mirror has a focal length of...Ch. 36 - A spherical mirror is to be used to form an image...Ch. 36 - Review. 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Locate...Ch. 36 - An object located 32.0 cm in front of a lens forms...Ch. 36 - An object is located 20.0 cm to the left of a...Ch. 36 - The projection lens in a certain slide projector...Ch. 36 - An objects distance from a converging lens is 5.00...Ch. 36 - A contact lens is made of plastic with an index of...Ch. 36 - A converging lens has a focal length of 10.0 cm....Ch. 36 - A converging lens has a focal length of 10.0 cm....Ch. 36 - A diverging lens has a focal length of magnitude...Ch. 36 - Prob. 36.47PCh. 36 - Suppose an object has thickness dp so that it...Ch. 36 - The left face of a biconvex lens has a radius of...Ch. 36 - In Figure P35.30, a thin converging lens of focal...Ch. 36 - An antelope is at a distance of 20.0 m from a...Ch. 36 - Prob. 36.52PCh. 36 - A 1.00-cm-high object is placed 4.00 cm to the...Ch. 36 - The magnitudes of the radii of curvature are 32.5...Ch. 36 - Two rays traveling parallel to the principal axis...Ch. 36 - Prob. 36.56PCh. 36 - Figure 35.34 diagrams a cross section of a camera....Ch. 36 - Josh cannot see objects clearly beyond 25.0 cm...Ch. 36 - Prob. 36.59PCh. 36 - A person sees clearly wearing eyeglasses that have...Ch. 36 - Prob. 36.61PCh. 36 - A certain childs near point is 10.0 cm; her far...Ch. 36 - A person is to be fitted with bifocals. She can...Ch. 36 - A simple model of the human eye ignores its lens...Ch. 36 - A patient has a near point of 45.0 cm and far...Ch. 36 - A lens that has a focal length of 5.00 cm is used...Ch. 36 - The distance between the eyepiece and the...Ch. 36 - The refracting telescope at the Yerkes Observatory...Ch. 36 - A certain telescope has an objective mirror with...Ch. 36 - Astronomers often take photographs with the...Ch. 36 - Prob. 36.71APCh. 36 - A real object is located at the zero end of a...Ch. 36 - The distance between an object and its upright...Ch. 36 - Prob. 36.74APCh. 36 - Andy decides to use an old pair of eyeglasses to...Ch. 36 - Prob. 36.76APCh. 36 - The lens and mirror in Figure P36.77 are separated...Ch. 36 - Two converging lenses having focal lengths of f1 =...Ch. 36 - Figure P36.79 shows a piece of glass with index of...Ch. 36 - Prob. 36.80APCh. 36 - The object in Figure P36.81 is midway between the...Ch. 36 - In many applications, it is necessary to expand or...Ch. 36 - Prob. 36.83APCh. 36 - Prob. 36.84APCh. 36 - Two lenses made of kinds of glass having different...Ch. 36 - Why is the following situation impossible?...Ch. 36 - An object is placed 12.0 cm to the left of a...Ch. 36 - An object is placed a distance p to the left of a...Ch. 36 - An observer to the right of the mirror-lens...Ch. 36 - In a darkened room, a burning candle is placed...Ch. 36 - Prob. 36.91APCh. 36 - An object 2.00 cm high is placed 40.0 cm to the...Ch. 36 - Assume the intensity of sunlight is 1.00 kW/m2 at...Ch. 36 - A zoom lens system is a combination of lenses that...Ch. 36 - Figure P36.95 shows a thin converging lens for...Ch. 36 - A floating strawberry illusion is achieved with...Ch. 36 - Consider the lensmirror arrangement shown in...
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