Zoom Lens: Consider the simple model of the zoom lens as shown in (A). The converging lens has focal length f₁ = 12 cm, and the diverging lens has focal length f₂ = -12 cm. The lenses are separated by 4 cm. (a) For a distant object, where is the image of the converging lens? (b) The image of the converging lens serves as the object for the diverging lens. What is the object distance for the diverging lens? (c) Where is the final image? (d) Repeat parts (a), (b), and (c) for the situation shown in (B), in which the lenses are separated by 8 cm.

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

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

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

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

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

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

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