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 64E

(a)

To determine

The focal length of lenses she required.

(a)

Expert Solution
Check Mark

Answer to Problem 64E

The focal length lenses she required is 0.5m.

Explanation of Solution

Write the expression for the focal length of the lenses.

  1f=1v+1u        (I)

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

Conclusion:

Substitute 0.5m for v and for u in equation (I).

  1f=1(0.5m)+1=1(0.5m)+0=2m1

Simplify for the focal length.

f=12m=0.5m

Thus, the focal length lenses she required is 0.5m.

(b)

To determine

The near point without the glasses.

(b)

Expert Solution
Check Mark

Answer to Problem 64E

The near point without the glasses is 0.167m.

Explanation of Solution

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

  Pf=1xf+1D        (II)

Here, Pf is the power of eye at far point, xf is the far point of the eye and D is the image distance of the eye.

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

  Pn=Pf+A        (III)

Here, Pn is the power of the eye at near point and A is power of accommodation of the eye.

Write the expression for the near point of the eye.

  xn=DD(Pn)1        (IV)

Here, xn is the near point.

Conclusion:

Substitute 0.5m for xf and 0.02m for D in equation (II).

  Pf=1(0.5m)+1(0.02m)=(2+50)diaptors=52diaptors

Substitute 52diaptors for Pf and 4diaptors for A in equation (III).

  Pn=(52diaptors)+(4diaptors)=56diaptors

Substitute 56diaptors for Pn and 0.02m for D in equation (IV).

  xn=(0.02m)(0.02m)(56diaptors)1=0.167m

The near point without the glasses is 0.167m.

(c)

To determine

The near point with the glasses.

(c)

Expert Solution
Check Mark

Answer to Problem 64E

The near point with the glasses is 0.25m.

Explanation of Solution

Write the expression for the power of eye at infinity.

  Pf=1+1D        (V)

Here, Pf is the power of eye at infinity and D is the image distance of the eye.

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

  Pn=Pf+A        (VI)

Here, Pn is the power of the eye at near point with the glasses and A is power of accommodation of the eye.

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

  xn=DD(Pn)1        (VII)

Here, xn is the near point.

Conclusion:

Substitute 0.02m for D in equation (V).

  Pf=0+1(0.02m)=50diaptors

Substitute 50diaptors for Pf and 4diaptors for A in equation (VI).

  Pn=(50diaptors)+(4diaptors)=54diaptors

Substitute 54diaptors for Pn and 0.02m for D in equation (IV).

  xn=(0.02m)(0.02m)(54diaptors)1=0.25m

The near point with the glasses is 0.25m.

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Chapter 24 Solutions

General Physics, 2nd Edition

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