Question

Bill is farsighted and has a near point located 125 cm from his eyes. Anne is also farsighted, but her near point is 75.0 cm from her eyes. Both have glasses that correct their vision to a normal near point (25.0 cm from the eyes), and both wear the glasses 2.0 cm from the eyes. Relative to the eyes, what is the closest object that can be seen clearly (a) by Anne when she wears Bill's glasses and (b) by Bill when he wears Anne's glasses

Answer 1

Answer:

a) The closest object Anne can see, while wearing Bill's glasses is at approximately 22.39 cm relative to her eyes

b) The closest object Bill can see while wearing Anne's glasses is at approximately 26.38 cm relative to his eyes

Explanation:

The point where Bill has a near point = 125 cm

The point where Anne has a near point = 75.0 cm

Their vision are both corrected to the normal near point = 25.0 cm

The distance of their glasses from the eye, d = 2.0 cm

The lens formula is presented as follows;

$\dfrac{1}{f} =\dfrac{1}{d_0} +\dfrac{1}{d_i}$

The required distance of the of the object from Bill's glass, $d_{oB}$ = 25 cm - 2.0 cm = 23.0 cm

The distance of the of the image from Bill's glass, $d_{iB}$ = -(125 cm - 2.0 cm) = -123 cm

Therefore, for Bill, we have;

$\dfrac{1}{f_B} =\dfrac{1}{d_{0B}} +\dfrac{1}{d_{iB}}$

Plugging in the values gives;

$\dfrac{1}{f_B} =\dfrac{1}{23} -\dfrac{1}{123} = \dfrac{100}{2,829}$

Therefore;

$f_B$ = 2,829/100 cm = 28.29 cm

The required distance of the of the object from Anne's glass, $d_{oA}$ = 25 cm - 2.0 cm = 23.0 cm

The distance of the of the image from Anne's glass, $d_{iA}$ = -(75 cm - 2 cm) = -73 cm

The focal length for Anne is therefore;

$\dfrac{1}{f_A} =\dfrac{1}{d_{0A}} +\dfrac{1}{d_{iA}}$

Plugging in the values gives;

$\dfrac{1}{f_A} =\dfrac{1}{23} -\dfrac{1}{73} = \dfrac{50}{1,679}$

Therefore, we have;

$f_A$ = 1,679/50 cm = 33.58 cm

a) When Anne wears Bill's lasses, we have;

$\dfrac{1}{f_B} =\dfrac{1}{d_{0A}} +\dfrac{1}{d_{iA}}$

Therefore, we get;

$\dfrac{1}{28.29} =\dfrac{1}{d_{0A}} -\dfrac{1}{73}$

$\dfrac{1}{d_{0A}} = \dfrac{1}{28.29} + \dfrac{1}{73} = \dfrac{10,129}{206,517}$

$d_{0A}$ ≈ 20.39 cm

The distance of the closest object Anne can see, from her eye, $d_{oe}$ =  $d_{0A}$ + d

∴ $d_{oe}$ ≈ 20.39 cm + 2.0 cm = 22.39 cm

The distance of the closest object Anne can see, from her eye, $d_{oe}$ ≈ 22.39 cm

b) The closest object that can be seen when Bill wears Anne's glasses, we have;

$\dfrac{1}{f_A} =\dfrac{1}{d_{0B}} +\dfrac{1}{d_{iB}}$

Therefore;

$\dfrac{1}{33.58} =\dfrac{1}{d_{0B}} -\dfrac{1}{123}$

$\dfrac{1}{d_{0B}} = \dfrac{1}{33.58} +\dfrac{1}{123} = \dfrac{7,829}{206,517}$

∴ $d_{oB}$ ≈ 26.38 cm

The closest object Bill can see while wearing Anne's glasses,  $d_{oB}$ ≈ 26.38 cm.