Question

The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.54 dp) is 27.5 cm. What is the near distance?(Assume a distance between the eyeglasses and the eyes to be 2.50 cm) please show work clearly and explain. i am used to variables Di and Do if you use other variables like s or v or p, PLEASE explain what those variables represent! thank u!

Answer 1

Given:

• Power of eyeglasses ($P$) = 1.54 D

• Distance from eyeglasses to book ($D_o$) = 27.5 cm

• Distance from eyeglasses to eyes ($d$) = 2.50 cm

• 
  

Goal:

Find the near distance ($D_{\text{near}}$), which is the closest distance the eye can focus without the glasses.

Variables:

• $D_o$: Object distance (book to glasses) = 27.5 cm

• $D_i$: Image distance (glasses to eye's near point)

• $D_{\text{near}}$: Near point distance (eye's natural closest focus point)

• $d$: Distance from glasses to eye = 2.50 cm

Steps:

1. Convert distances to meters (since Power is in diopters, $D$):

• $D_o=27.5\text{cm}=0.275\text{m}$

• $d=2.50\text{cm}=0.025\text{m}$

2. Use the lens formula to find the image distance ($D_i$):

   $$P=\frac{1}{D_i}-\frac{1}{D_o}$$

   Rearrange to solve for $D_i$:

   $$\frac{1}{D_i}=P+\frac{1}{D_o}=1.54+\frac{1}{0.275}=1.54+3.636=5.176\text{D}$$$$D_i=\frac{1}{5.176}\approx 0.1932\text{m}=19.32\text{cm}$$

3. Relate the image distance to the near point ($D_{\text{near}}$):
   The image formed by the glasses acts as the object for the eye. Since the eye is 2.50 cm from the glasses:

   $$D_{\text{near}}=D_i+d=19.32\text{cm}+2.50\text{cm}=21.82\text{cm}$$

Answer:

The near distance (closest distance the eye can focus without glasses) is 21.8 cm.

Answer 2