91Ó°ÊÓ

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.

Focal length of lens,

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

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

By using lens formula

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

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

Focal length of lens,

$$\begin{gathered} ⇒Power,P=\frac{1}{Focallength,f} \\ ⇒f=\frac{1}{P}=\frac{100}{-1.30}cm=-77cm \end{gathered}$$

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

By using lens formula

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

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