General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 24, Problem 66E

(a)

To determine

The far point of the man.

(a)

Expert Solution
Check Mark

Answer to Problem 66E

The far point of the man is 0.0714m.

Explanation of Solution

Write the expression for the power of eye at near point.

  Pn=1xn+1D        (I)

Here, Pn is the power of eye at near point, xn is the near point of the eye and D is the image distance of the eye.

Write the expression for the power of the eye at far point.

  Pf=Pn+A        (II)

Here, Pf is the power of the eye at far point and A is power of accommodation of the eye.

Write the expression for the power of the eye at far point.

  Pf=1xf+1D        (III)

Here, xf is the far point of the eye.

Conclusion:

Substitute 0.1m for xn and 0.02m for D in equation (I).

  Pn=1(0.1m)+1(0.02m)=(10+50)diaptors=60diaptors

Substitute 60diaptors for Pn and 4diaptors for A equation (II).

  Pf=60diaptors4diaptors=56diaptors

Substitute 56diaptors for Pf and 0.02m for D in equation (III).

  56diaptors=1xf+1(0.02m)56diaptors=1xf+50diaptors

Simplify the above expression as:

  1xf=56diaptors50diaptors=6diaptors

Simplify the above expression for the far point as:

  xf=16m=0.1667m

Thus, the far point of the man is 0.1667m.

(b)

To determine

The power of lens he needed.

(b)

Expert Solution
Check Mark

Answer to Problem 66E

The power of lens he needed is 60diaptors.

Explanation of Solution

Write the expression for the power of eye at near point.

  Pn=1xn+1D        (I)

Here, Pn is the power of eye at near point, xn is the near point of the eye and D is the image distance of the eye.

Conclusion:

Substitute 0.1m for xn and 0.02m for D in equation (i).

  Pn=1(0.1m)+1(0.02m)=(10+50)diaptors=60diaptors

Thus, the power of lens he needed is 60diaptors.

(c)

To determine

The near point with the glasses.

(c)

Expert Solution
Check Mark

Answer to Problem 66E

The near point with the glasses is 0.1m.

Explanation of Solution

Write the expression for the power of eye at near point.

  Pn=1xn+1D        (II)

Here, Pn is the power of glasses, xn is the near point of the eye with the glasses and D is the image distance of the eye.

Conclusion:

Substitute 0.1m for xn and 0.02m for D in equation (II).

  60diaptors=1xn+1(0.02m)

Simplify the above expression for the near point.

  xn=1(10+50)diaptors=0.1m

Thus, the near point with the glasses is 0.1m.

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Chapter 24 Solutions

General Physics, 2nd Edition

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