A person whose near-point distance is 45.5 cm wears a pair of glasses that are 2.1 cm from her eyes. With the aid of these glasses, she can now focus on objects 25 cm away from her eyes. Required:
a. Find the focal length of her glasses.
b. Find the refractive power of her glasses.

a

b

Explanation:

From the question we are told that

   The  near point distance is  

    The  distance of the glasses from the eye is  

    The  distance of an object she can focus with the glass is  

Generally the image distance is mathematically evaluated  as

Generally the object distance is mathematically represented as

The  negative sign tells us that the image was formed behind the eye  

Generally the lens formula  is mathematically represented as  

=>  

=>  

 converting to meters  

Thus the refractive power is mathematically represented as

=>  

=>  

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day