EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 32, Problem 27P

(a)

To determine

The image distances for given object distances.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The object distances are, 55cm , 24cm , 12cm , 8.0cm and 1.0cm .

The radius of curvature is 24cm .

Formula used:

Write expression for mirror equation.

  1v+1u=1f

Here, v is image distance, u is object distance and f is focal length.

Substitute r2 for f in above expression.

  1v+1u=2r

Solve above expression for v .

  v=ru2ur  ....... (1)

Calculation:

Substitute v1 for v , 24cm for r and 55cm for u in equation (1).

  v1=(55cm)(24cm)2(55cm)(24cm)v1=9.85cm

Substitute v2 for v , 24cm for r and 24cm for u in equation (1).

  v2=(24cm)(24cm)2(24cm)(24cm)v2=8cm

Substitute v3 for v , 24cm for r and 12cm for u in equation (1).

  v3=(24cm)(12cm)2(12cm)(24cm)v3=6cm

Substitute v4 for v , 24cm for r and 8cm for u in equation (1).

  v4=(24cm)(8cm)2(8cm)(24cm)v4=4.8cm

Substitute v5 for v , 24cm for r and 1.0cm for u in equation (1).

  v5=(24cm)(1cm)2(1cm)(24cm)v5=0.92cm

Conclusion:

Thus, the image distance is 9.85cm for 55cm ,the image formation for object at 24cm is 8cm , the image formation for object at 12cm is 6cm , the image formation for object at 8cm is 4.8cm and the image formation for object at 1.0cm is at 0.92cm .

(b)

To determine

The magnification of each given object distance.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The object distances are, 55cm , 24cm , 12cm , 8.0cm and 1.0cm .

The radius of curvature is 24cm .

Formula used:

Write expression for magnification for mirror.

  m=vu  ....... (1)

Calculation:

Substitute m1 for m , 55cm for u and 9.85cm for v in equation (1).

  m1= (9.85cm)55m1=0.18

Substitute m2 for m , 24cm for u and 8cm for v in equation (1).

  m2=(8cm)24cmm2=0.33

Substitute m3 for m , 12cm for u and 6cm for v in equation (1).

  m3=(6cm)12m3=0.5

Substitute m4 for m , 8.0cm for u and 4.8cm for v in equation (1).

  m4=(4.8cm)8.0m4=0.60

Substitute m5 for m , 1.0cm for u and 0.92cm for v in equation (1).

  m5=(0.92)1.0m5=0.92

Conclusion:

Thus, the magnification for first case is 0.18 , for second case is 0.33 , for third case is 0.50 , for fourth case is 0.60 and for fifth case is 0.92 .

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Chapter 32 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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