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 75E

(a)

To determine

The far point of the man without glasses.

(a)

Expert Solution
Check Mark

Answer to Problem 75E

The far point of the man without glasses is 0.125m.

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=PnA        (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=10.1m+10.02m=(10+50)diaptors=60diaptors

Substitute 60diaptors for Pn and 2diaptors for A in equation (II).

  Pf=60diaptors2diaptors=58diaptors

Substitute 58diaptors for  Pf and 0.02m for D in equation (III).

  58diaptors=1xf+1(0.02m)58diaptors=1xf+50diaptors

Simplify the above expression.

  1xf=58diaptors50diaptors=8diaptors

Simplify the above expression for the far point.

  xf=18m=0.125m

Thus, the far point of the man without glasses is 0.125m.

(b)

To determine

The far point of the man with the correct glasses.

(b)

Expert Solution
Check Mark

Answer to Problem 75E

The far point of the man with the correct glasses is 0.5m.

Explanation of Solution

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

  Pn=1xn+1D        (IV)

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

Write the expression for the power of lens in the glasses.

  Pf=PnA        (V)

Write the expression for the far point.

  Pf=1xf+1D        (VI)

Conclusion:

Substitute 0.25m for xn and 0.02m for D in equation (IV).

  Pn=1(0.25m)+1(0.02m)=(4+50)diaptors=54diaptors

Substitute 54diaptors for Pn and 2diaptors for A equation (V).

  Pf=54diaptors2diaptors=52diaptors

Substitute 52diaptors for  Pf and 0.02m for D in equation (VI).

  52diaptors=1xf+1(0.02m)52diaptors=1xf+50diaptors

Simplify the above expression.

  1xf=52diaptors50diaptors=2diaptors

Simplify the above expression for the far point as:

  xf=12m=0.5m

Thus, the far point of the man with the correct glasses is 0.5m.

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Chapter 24 Solutions

General Physics, 2nd Edition

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