SATHEE: Optics 7 Question 18

Optics 7 Question 18

18. Two thin convex lenses of focal lengths $f_{1}$ and $f_{2}$ are separated by a horizontal distance $d$ (where $d < f_{1}, d < f_{2}$ ) and their centres are displaced by a vertical separation $Δ$ as shown in the figure.

(1993)

Taking the origin of coordinates, $O$ , at the centre of the first lens, the $x$ and $y$ -coordinates of the focal point of this lens system, for a parallel beam of rays coming from the left, are given by

(1993; 2M)

(a) $x = \frac{f_{1} f_{2}}{f_{1} + f_{2}}, y = Δ$

(b) $x = \frac{f_{1} (f_{2} + d)}{f_{1} + f_{2} - d}, y = \frac{Δ}{f_{1} + f_{2}}$

(d) $x = \frac{f_{1} f_{2} + d (f_{1} - d)}{f_{1} + f_{2} - d}, y = 0$

Numerical Value

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Solution:

From the first lens parallel beam of light is focused at its focus i.e. at a distance $f_{1}$ from it. This image $I_{1}$ acts as virtual object for second lens $L_{2}$ . Therefore, for $L_{2}$

$\begin{aligned} u & = + (f_{1} - d), f = + f_{2} \\ ∴ & \frac{1}{v} & = \frac{1}{f} + \frac{1}{u} = \frac{1}{f_{2}} + \frac{1}{f_{1} - d} \\ Hence, & v & = \frac{f_{2} (f_{1} - d)}{f_{2} + f_{1} - d} \end{aligned}$

Therefore, $x$ -coordinate of its focal point will be

$\begin{aligned} x & = d + v = d + \frac{f_{2} (f_{1} - d)}{f_{2} + f_{1} - d} \\ = \frac{f_{1} f_{2} + d (f_{1} - d)}{f_{1} + f_{2} - d} \end{aligned}$

Linear magnification for $L_{2}$

$\begin{aligned} m & = \frac{v}{u} = \frac{f_{2} (f_{1} - d)}{f_{2} + f_{1} - d} \cdot \frac{1}{f_{1} - d} \\ = \frac{f_{2}}{f_{2} + f_{1} - d} \end{aligned}$