Q51E - University Physics with Modern Physics

Question

(a) Where is the near point of an eye for which a contact lens with a power of +2.75 diopters is prescribed? (b) Where is the far point of an eye for which a contact lens with a power of -1.30 diopters is prescribed for distant vision?

Step-by-Step Solution

Answer

a) the near point of the eyes without lens is 80 cm .

b) the far point of the eyes without lens is 77cm .

Step 1: Basic definition

Thin Lens formula:

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

Here, 

u = Object distance from lens

v = Image distance from lens

f = Focal length of the lens

Sign Convention:

1. Object Distance (u): (+) in front of lens; (-) in the back of lens
2. Image Distance (v): (-) in front of lens; (+) in the back of lens
3. Focal Length (f): (+) for convergent (convex) lens; (-) for divergent (concave) lens

Power of a lens is defined as the reciprocal of its focal length. Thus, converging or convex lens have positive powers and diverging or concave lens have negative powers.

Step 2: Near point of the eye

Focal length of lens,

 

$$\begin{gathered} \Rightarrow Power,p=\frac{1}{focallength,f} \\ \Rightarrow f=\frac{1}{P}=\frac{100}{+2.75}cm \\ =36.36cm \end{gathered}$$

 

Object distance, u = +25.0 cm (Near point of eyes with lens)

 

By using lens formula

 

$$\begin{gathered} \Rightarrow \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\ \Rightarrow \frac{1}{+25}+\frac{1}{v}=\frac{1}{36.36} \\ \Rightarrow \frac{1}{v}=\frac{1}{+36.36}-\frac{1}{+25} \\ \Rightarrow \frac{1}{v_1}=\frac{25-36.36}{36.36*25} \\ \Rightarrow v=-80cm \end{gathered}$$

 

Therefore, the near point of the eyes without lens is  80 cm

Step 3: Far point of the eye

Focal length of lens,

 

$$\begin{gathered} \Rightarrow Power,P=\frac{1}{Focal length,f} \\ \Rightarrow f=\frac{1}{P}=\frac{100}{-1.30}cm=-77cm \end{gathered}$$

 

Object distance, $u=\infty$ (Far point of eyes with lens)

 

By using lens formula

 

$$\begin{gathered} \Rightarrow \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\ \Rightarrow \frac{1}{\infty }+\frac{1}{v}=\frac{1}{-77} \\ \Rightarrow \frac{1}{v}=\frac{1}{-77}-\frac{1}{\infty } \\ \Rightarrow v=-77cm \end{gathered}$$

 

Therefore, the far point of the eyes without lens is 77cm .