91Ó°ÊÓ

An optical lens is a transparent transmissive optical component that is used to either converge or diverge the light emitting from a object. These transmitted light forms an image of that object.

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

Location and Height of I₁

Focal length of 1^st lens, f₁ = +12 cm

Object distance, u = +∞

By using lens formula

$$\begin{gathered} ⇒\frac{1}{u}+\frac{1}{v_1}=\frac{1}{f_1} \\ ⇒\frac{1}{∞}+\frac{1}{v_1}=\frac{1}{+12} \\ ⇒\frac{1}{v_1}=\frac{1}{+12} \\ ⇒v_1=+12cm \end{gathered}$$

For a distance object, converging lens forms an image at 12 cm from its optical center.

Location and Height of I₂

Focal length of 2^nd lens, f₂ = -12 cm

The image of the converging lens serves as the object for diverging lens and its object distance is u = -(12 cm – 4) cm = -8 cm

By using lens formula

role="math" localid="1663930036918" $$\begin{gathered} ⇒\frac{1}{u}+\frac{1}{v^2}=\frac{1}{f_2} \\ ⇒\frac{1}{-8}+\frac{1}{v_2}=\frac{1}{-12} \\ ⇒\frac{1}{v_2}=-\frac{1}{12}+8 \\ ⇒\frac{1}{v_2}=\frac{-2+3}{24} \\ ⇒v_2=24cm \end{gathered}$$

Therefore, the final image forms at 24 cm to the right of the diverging lens.

Location and Height of I₁

Focal length of 1^st lens, f₁ = +12 cm

Object distance, u = +∞

By using lens formula

$$\begin{gathered} ⇒\frac{1}{u}+\frac{1}{v_1}=\frac{1}{f_1} \\ ⇒\frac{1}{∞}+\frac{1}{v_1}=\frac{1}{+12} \\ ⇒\frac{1}{v_1}=\frac{1}{+12} \\ ⇒v_1=+12cm \end{gathered}$$

For a distance object, converging lens forms an image at 12 cm from its optical center.

Location and Height of I₂

Focal length of 2^nd lens, f₂ = -12 cm

The image of the converging lens serves as the object for diverging lens and its object distance is u = -(12 cm – 8) cm = -4 cm

By using lens formula

role="math" localid="1663930135448" $$\begin{gathered} ⇒\frac{1}{u}+\frac{1}{v^2}=\frac{1}{f_2} \\ ⇒\frac{1}{-4}+\frac{1}{v_2}=\frac{1}{-12} \\ ⇒\frac{1}{v_2}=-\frac{1}{12}+\frac{1}{4} \\ ⇒\frac{1}{v_2}=\frac{-1+3}{12} \\ ⇒v_2=+6cm \end{gathered}$$

Therefore, the final image forms at 6 cm to the right of the diverging lens.