Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
5th Edition
ISBN: 9781305586871
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 26, Problem 59P

(a)

To determine

Intensity of light at the surface of the bulb.

(a)

Expert Solution
Check Mark

Answer to Problem 59P

The intensity is 1.4kW/m2_.

Explanation of Solution

Write the equation for intensity

    I=PA        (I)

Here I is the intensity, P is the power and A is the area

Write the equation for surface area.

    A=4π(d2)2        (II)

Here d is the diameter of the bulb.

Substitute (II) in (I)

    I=P4π(d2)2=Pπd2        (III)

Conclusion:

Substitute 4.5W for P, 3.2cm for d in(III)

    I=4.5Wπ(3.2cm(102m1cm))2=4.5Wπ(3.2×102m)2=1.4kW/m2

The intensity is 1.4kW/m2_.

(b)

To determine

Light intensity at 7.2cm away from the center.

(b)

Expert Solution
Check Mark

Answer to Problem 59P

The intensity is 6.91mW/m2_.

Explanation of Solution

Write the equation for radius

    r=d2        (IV)

Here d is the diameter and r is the radius

Substitute (IV) in (III)

    I=P4πr2        (V)

Conclusion:

Substitute 4.5W for P, 7.2m for r in (V)

    I=4.5W4π(7.2m)2=6.91mW/m2

The intensity is 6.91mW/m2_.

(c)

To determine

Diameter of the bulb’s image.

(c)

Expert Solution
Check Mark

Answer to Problem 59P

Image diameter is 0.164cm_.

Explanation of Solution

Write the mirror equation

      1f=1p+1q        (VI)

Here p is the object distance, q is the image distance and f is the focal length.

Rearrange (VI) in terms of q

    q=pfpf        (VII)

Diameter of the image is same as that of the height of the image.

Write the equation for image magnification

  M=qp        (VIII)

Here M is the image magnification.

Write the equation for magnification in terms of image height.

    M=hh        (IX)

Here h' is the image height and h is the object height.

Equate (VIII) and (IX) and write in terms of h'

    h=qph        (X)

Conclusion:

Substitute 7.2m for p and 0.35m for f in (VII)

    q=(7.2m)(0.35m)7.2m0.35m=0.368m

Substitute 0.389m for q, 7.2m for p and 3.2cm for h in (X)

    h=0.389m7.2m(3.2cm(102m1cm))=0.389m7.2m(3.2×102m)=0.164cm

Image diameter is 0.164cm_.

(d)

To determine

Light intensity at the image.

(d)

Expert Solution
Check Mark

Answer to Problem 59P

The intensity is 58.1W/m2_.

Explanation of Solution

Write the equation for area of the bulb

    A=π(d2)2        (XI)

Here d is the diameter.

Substitute (XI) in (I)

    I=Pπ(d2)2        (XII)

Conclusion:

Substitute 6.91mW/m2 for P, 15cm for d in (XII)

    I=6.91mW/m2π(15cm(102m1cm)2)2=6.91mW/m2π(15×102m2)2=58.1W/m2

The intensity is 58.1W/m2_.

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Chapter 26 Solutions

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)

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