Question

The distance between any two sources of light is 24 cm. Determine where a convergent lens of focal length 9 cm be placed so that the image of both sources be formed at one point.

Hint:

Put the lens at distance \(x\) from one source and \(24-x\) from the other, apply the lens formula to each, and set the images to coincide (\(v_1=-v_2\)). You get \(x=6\) cm or 18 cm.

Answer 1

Step 1: Set up the geometry.
The two point sources \(S_1\) and \(S_2\) are 24 cm apart on the principal axis, and the lens is placed between them. Let the lens be at a distance \(x\) from \(S_1\); then it is at \((24 - x)\) from \(S_2\). Light from \(S_1\) travels one way through the lens and light from \(S_2\) travels the opposite way. For both images to fall on one point, the two images must be formed at the same place.

Step 2: Image of the first source.
Using the lens formula \(\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\) with \(u=-x\), \(f=9\):
\[ \frac{1}{v_1} = \frac{1}{9} - \frac{1}{x} \quad\Rightarrow\quad v_1 = \frac{9x}{x-9} \]

Step 3: Image of the second source.
For \(S_2\), object distance is \((24-x)\):
\[ v_2 = \frac{9(24-x)}{(24-x)-9} = \frac{9(24-x)}{15-x} \]

Step 4: Condition that both images coincide.
The two images are formed on opposite sides of the lens, so they meet at one common point when \(v_1 = -v_2\):
\[ \frac{9x}{x-9} = -\frac{9(24-x)}{15-x} \]
\[ x(15-x) = -(24-x)(x-9) \]
Expanding: \(15x - x^2 = -(33x - x^2 - 216) = x^2 - 33x + 216\).
\[ 15x - x^2 = x^2 - 33x + 216 \]
\[ 2x^2 - 48x + 216 = 0 \quad\Rightarrow\quad x^2 - 24x + 108 = 0 \]

Step 5: Solve the quadratic.
\[ x = \frac{24 \pm \sqrt{576 - 432}}{2} = \frac{24 \pm \sqrt{144}}{2} = \frac{24 \pm 12}{2} \]
\[ x = 18\ \text{cm} \quad\text{or}\quad x = 6\ \text{cm} \]
Both roots describe the same physical position: the lens must be placed 6 cm from one source (and hence 18 cm from the other). Check at \(x = 6\): \(v_1 = \dfrac{9\times6}{6-9} = -18\) cm and \(v_2 = \dfrac{9\times18}{15-6} = 18\) cm, so both images form at a common point 18 cm from the lens.

\[\boxed{\text{Place the lens 6 cm from one source (18 cm from the other)}}\]