Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 24, Problem 63P

(a)

To determine

The uncorrected far point.

(a)

Expert Solution
Check Mark

Answer to Problem 63P

The uncorrected far point is 1.602m.

Explanation of Solution

Write the equation to find the distance between the glass and book.

  p=dbookdglass                                                                                                       (I)

Here, p is the distance between the glass and book, dbook is the distance from relaxed eye to book, and dglass is the separation between glass and eye.

Write the lens formula.

  1p+1q=1f                                                                                                              (II)

Here, q is the uncorrected far point, p is the object distance, f is the focal length, and 1f is the power.

Rewrite the above relation in terms of q.

    q=(1f1p)1                                                                                                          (III)

Write the equation to find the corrected far point.

    q=q+dglass                                                                                                           (IV)

Here, q is the corrected far point.

Conclusion:

Substitute 40.0cm for dbook and 2.0cm for dglass in equation (I) to find p.

    p=40.0cm2.0cm=38.0cm

Substitute 2.0D for 1f and 38.0cm for p in equation (III) to find q.

  q=(2.0D1(38.0cm(102m1cm)))1=10.625m1=1.6m

Substitute 1.6m for q and 2.0cm for dglass in equation (IV) to find q.

    q=1.6m+2.0cm(102m1cm)=1.602m

Therefore, the uncorrected far point is 1.602m.

(b)

To determine

Power of lens required for distance vision.

(b)

Expert Solution
Check Mark

Answer to Problem 63P

Power of lens required for distance vision is 0.63D.

Explanation of Solution

Object distance is taken as infinity for distant objects.

Write the equation to find q in the given situation.

  q=q+dglass                                                                                                           (V)

Write the relation between f and power of lens.

    P=1f                                                                                                                       (VI)

Here,P is the power of lens.

Conclusion:

Substitute 1.602 for q and 2.0cm for dglass in equation (V) to find q.

  q=1.602m+2.0cm(102m1cm)=1.582m

Substitute for p and 1.582m for q in equation (II) to find 1f.

    1+11.582m=1f00.63m1=1f

Substitute 0.63m1 for 1f in equation (VI) to find P.

  P=0.63m1(1D1m1)=0.63D

Therefore, the power of lens required for distance vision is 0.63D.

(c)

To determine

Power of two lenses required for clear vision from 25cm to infinity.

(c)

Expert Solution
Check Mark

Answer to Problem 63P

Power of two lenses required for clear vision from 25cm to infinity are 0.63D and 3.3D.

Explanation of Solution

Object distance is taken as infinity for distant objects.

Write the equation to find p.

    p=ddglass                                                                                                  (VII)

Here, d is the near point of eye.

Write the equation to find q.

  q=dun+dglass                                                                                         (VIII)

Conclusion:

Substitute 25cm for d and 2.0cm for dglass in equation (VII) to find p.

  p=25cm2.0cm=23cm(102m1cm)=0.23m

Substitute 1.0m for d and 2.0cm for dglass in equation (VIII) to find q.

    q=1.0m+2.0cm(102m1cm)=0.98m

Substitute 0.23m for p and 0.98m for q in the lens formula to find 1f.

    10.23m+10.98m=1f3.3m1=1f

Substitute 3.3m1 for 1f in equation (VI) to find P.

  P=3.3m1(1D1m1)=3.3D

Therefore, Power of two lenses required for clear vision from 25cm to infinity are 0.63D and 3.3D.

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Chapter 24 Solutions

Physics

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