Question

What do you mean by an equivalent lens? The system of two equivalent lenses are placed at a distance then derive the equation of equivalent focus and power.

Answer 1

Hint: When two or more than two lenses are placed in a common axis, the lenses are said to be co-axial. Use this information to arrive at the definition of equivalent lenses. To find the equivalent focus and power use the formula of two co-axial lenses placed at a distance and equate the focal lengths.

Complete step by step answer:
Let us consider two lenses of two different focal lengths ${f_1}$ and ${f_2}$ placed coaxially (in one common axis). If we can find one single lens that can produce the same image at the same place and of the same size as the above setup, then we can state that this one single lens is equivalent to the above combination of lenses. This is what an equivalent lens means.

Let us consider two converging lenses $1$ and $2$ , of second focal lengths ${f_1}$ and ${f_2}$ respectively, and keep them separated by a distance $d$ as shown in the given figure. A ray $AB$ moving parallel to the common axis of the lens is incident at a height $y_1$ above the axis of the first lens. This ray after refraction through the first lens is directed towards the second principle focal point $Q_1$ of this lens. This deviation $a_1$ produced by the first lens is given by:

$a_1 \approx \dfrac{{y_1}}{{{f_1}}}$

where ${f_1} = O_1Q_1$ is the second focal length of the first lens.

The emergent ray from the first lens is incident on the second lens at a height $y_2$ above the axis. After refraction through the second lens, the ray finally meets the principle axis at $F_2$ , which is the second focal point of the combination. The deviation $a_2$ produced by the second lens is given by

$a_2 \approx \dfrac{{y_2}}{{{f_2}}}$

where ${f_2}$ is the second focal length of the second lens.

The conjugate rays $AB$ and $CF_2$ when produced intersect at $E_2$ . if now a converging lens be placed at $E_2H_2$ having its second focus at $F_2$ and focal length $F_2 = H_2F_2$ , then the parallel ray $ABE_2$ incident at $E_2$ of the lens $E_2H_2$ will be refracted along $E_2CF_2$ producing the same total deviation $a$ as that produced by the two lenses together. Hence the lens of focal length $H_2F_2$ placed at $E_2H_2$ will be equivalent to the lens combination. The plane $E_2H_2$ is called the second principle plane and the point $H_2$ , the second principle point. The deviation produced by the equivalent lens is given by

$a = \dfrac{{y_1}}{f}$

Now, $a = a_1 + a_2$

$\Rightarrow \dfrac{{y_1}}{f} = \dfrac{{y_1}}{{{f_1}}} + \dfrac{{y_2}}{{{f_2}}}$

$ \Rightarrow \dfrac{{y_1}}{f} = \dfrac{{y_1}}{{{f_1}}} + \dfrac{{y_1 - (y_1 - y_2)}}{{{f_2}}}$ - equation $(1)$

From the figure, we can write

$y_1 - y_2 = CG = d.a_1 = d.\dfrac{{y_1}}{{{f_1}}}$

Substituting this in the equation $(1)$ we get,

$\dfrac{{y_1}}{f} = \dfrac{{y_1}}{{{f_1}}} + \dfrac{{y_1}}{{{f_2}}} - \dfrac{{dy_1}}{{{f_1}{f_2}}}$

$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{d}{{{f_1}{f_2}}}$

Hence the equivalent focal length,

$f = - \dfrac{{{f_1}{f_2}}}{{d - ({f_1} + {f_2})}}\,\,(m)$

Since power $(P) = \dfrac{1}{f}$

The equivalent power is,

$P = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{d}{{{f_1}{f_2}}}\,\,(D)$

Note: The unit of power is dioptres $(D)$ . Convex lenses have the focal length as positive, therefore they have positive power, whereas concave lenses have focal length negative, therefore they have negative power.