Q99P - University Physics with Modern Physics

Question

A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?

Step-by-Step Solution

Answer

a)The near point is at 30.9cm when he is wearing an old glass resting 2cm in front of the eye.

b)The near point is at 29.2cm when is wearing contact lenses.

Step 1:Relationship between object and image distance.

Object’s distance is s and image distance is s’.

The relationship is  $\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{ƒ}$

Step 2:Formula for calculating power.

The power is calculated by finding the reciprocal of the focal length.

  $P=\frac{1}{ƒ}$

Step 3: Calculation of near point when wearing glasses.

The power is given as  $P=+2.25D$

Therefore the focal length is the reciprocal of P.

 $$\begin{gathered} ƒ=\frac{1}{P} \\ =\frac{1}{2.25} \\ =0.444\mathit{m} \end{gathered}$$ 

The distance between the image and lens is,  $s\text{'}=85-2=83\mathit{c}\mathit{m}$

Using the relationship between the object and the image,

$$\begin{gathered} s=\frac{s\text{'}ƒ}{s-ƒ} \\ =\frac{-83*44.4}{-83-44} \\ =28.9 \end{gathered}$$ 

Therefore the distance between the image and the eye is,

$28.9+2=30.9\mathit{c}\mathit{m}$

Step 4: calculation of near point when wearing contact lenses .

The distance between the image and the eye will be 85

Using the relationship between the object and the image,

 $$\begin{gathered} s=\frac{s\text{'}ƒ}{s-ƒ} \\ =\frac{-83*44.4}{-83-44} \\ =29.2\mathit{c}\mathit{m} \end{gathered}$$