Compound Lenses - Practice Questions & MCQ

Combination of thin lens in contact -

Till now we have discussed single lens, but what happen when two or more than two lenses combined together. For this let us consider two lenses A and B of focal length f₁ and f₂ placed in contact with each other. Let the object be placed at a point O beyond the focus of the first lens A (See the figure).

The first lens produces an image at I₁. Now the image at I₁ is real but it serves as a virtual object for the second lens B. The lens B produce the final image at I. Since the lenses are thin, we assume the optical centres of the lenses to be coincident. Let this central point be denoted by P. 

So for the lens A - 

$
\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1}
$

Similarly for lens B -

$
\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}
$

Adding both the equation, we get -

$
\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2} \ldots
$

Now, let us assume two lens-system equivalent to a single lens of focal length $f$, we have -

$
\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \ldots
$

So from (1) and (2), we can conclude that the focal length after combination of two thin lenses, we get -

$
\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}
$

Similarly the by combining any number of thin lenses in contact. we can get -

$
\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3} \quad \ldots
$

In terms of Power, we can write this equation as -

$
P=P_1+P_2+P_3+\ldots
$
 

Here P is the net power of the lens combination. Note that the sum is an algebraic sum of individual powers, so the power may be positive or negative because it may be combination of both concave or convex lenses. So some of the terms on the right side may be positive (for convex lenses) and some negative (for concave lenses).

In terms of magnification we can write the net magnification as -

$
m=m_1 \cdot m_2 \cdot m_3 \ldots
$
 

As we know that in combination of lens the image of the first lens is the object for the second lens. So the magnification due to the combination of lens is multiplication of individual magnification. 

The combination of lenses is needed in lenses for cameras, microscopes, telescopes and other optical instruments.

Lenses at a distance -

When two thin lenses are separated by a distance,it is equivalent to a thick lens, but not equivalent to single thin lens. Now let us take a special case in which the object is placed at infinity, so the combination may be replaced by a single thin lens. We shall now derive the position and focal length of the equivalent lens in this special case. To start with, let us derive an expression for the angle of deviation of a ray when it passes through a lens. 

Let 0 be a point object on the principal axis of a lens. Let OA be a incident ray on the lens at a point A which is at a height h above the optical centre. It is deviated through an angle   and comes out along AI. It strikes the principal axis at I where the image is formed. 

Let $\angle A O P=\alpha$ and $\angle A I P=\beta$. By triangle $O A I$,

$
\text { By exterior angle property } \Rightarrow \delta=\alpha+\beta
$

If the hieght h is very small as compare to the PI and PO , then the $\alpha, \beta$ will be very small,
So we can write,

$
\begin{gathered}
\alpha=\tan \alpha=h / O P \text { and } \beta=\tan \beta=h / P I \\
\delta=\frac{h}{P O}+\frac{h}{P I}
\end{gathered}
$

Now by sign convention we can write that the $\mathrm{PO}=-\mathrm{u}$ and $\mathrm{PI}=\mathrm{v}$, So the above equation can be written as -

$
\begin{gathered}
\delta=h\left(\frac{1}{v}-\frac{1}{u}\right) \\
\delta=\frac{h}{f}
\end{gathered}
$

Now let us consider two thin lenses are placed coaxially at a separation d. The incident ray AB and the emergent ray $C D$ intersect at point named as $E$. The perpendicular from $E$ to the principal axis falls at $P$. The equivalent lens should be placed at this position P. A ray ABE going parallel to the principal axis will go through the equivalent lens and emerge along ECD. The angle of deviation is

$
\delta=\delta_1+\delta_2
$

(From the exterior angle property of the triangle BEC)

                                                                The focal length of the equivalent lens is F = PD,

Using above equation

$
\begin{gathered}
\delta_1=\frac{h_1}{f_1}, \quad \delta_2=\frac{h_2}{f_2} \text { and } \delta=\frac{h_1}{F} \\
\text { As } \delta=\delta_1+\delta_2 \\
\frac{h_1}{F}=\frac{h_1}{f_1}+\frac{h_2}{f_2}
\end{gathered}
$

Now,

$
\begin{gathered}
h_1-h_2=P_2 G-P_2 C=C G \\
=B G \tan \delta_1=B G \delta_1 \\
\text { or, } h_1-h_2=d \frac{h_1}{f_1}
\end{gathered}
$

Thus, by ,

$
\begin{aligned}
\frac{h_1}{F} & =\frac{h_1}{f_1}+\frac{h_1}{f_2}-\frac{d\left(h_1 / f_1\right)}{f_2} \\
\text { or, } \frac{1}{F} & =\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}
\end{aligned}
$

Position of the Equivalent Lens

$
\text { We have, } \begin{aligned}
& P P_2=E G \\
= & G C \cot \delta \\
& =\frac{h_1-h_2}{\tan \delta}
\end{aligned}
$
 

                                                        $\begin{aligned} & \qquad \begin{array}{l}h_1-h_2=\frac{d h_1}{f_1} \text {. Also, } \delta=\frac{h_1}{F} \text { so that } \\ P P_2=\left(\frac{d h_1}{f_1}\right)\left(\frac{F}{h_1}\right)=\frac{d F}{f_1}\end{array} \\ & \qquad \frac{d . F}{f_1} \text { behind the second lens. }\end{aligned}$

Note - Both the above relation are true only for the special case of parallel incident beam. If the object is at a finite distance, one should not use the above equations.