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Step-by-Step Solution
Step 1: Write down the known information
• A convex lens has focal length $f = 20\,\text{cm}.$
• The lens produces the same magnification $m = 2$ for two different object distances $x_{1}$ and $x_{2}$ (with $x_{1} > x_{2}$).
Step 2: Recall the lens formula and magnification relation
The lens formula is given by
$
\frac{1}{f} = \frac{1}{v} + \frac{1}{u},
$
where
• $f$ is the focal length of the lens,
• $u$ is the object distance (measured from the lens),
• $v$ is the image distance (measured from the lens).
The (absolute) lateral magnification $m$ is
$
m = \left|\frac{v}{u}\right|.
$
Step 3: Express magnification in terms of $u$ and $f$
Using the lens formula in the form $v = \frac{uf}{u-f}$ (for a convex lens), the magnification can be written as
$
m = \left|\frac{v}{u}\right| = \left| \frac{f}{u - f} \right|.
$
We are given $m = 2,$ so
$
2 = \left|\frac{f}{u - f}\right|.
$
Step 4: Solve for the two possible object distances
From
$
2 = \left|\frac{f}{u - f}\right|,
$
we get
$
\left|u - f\right| = \frac{f}{2}.
$
Hence there are two possibilities:
Case 1: $u - f = \frac{f}{2}$, which gives $u = \frac{3f}{2}.$
Case 2: $u - f = - \frac{f}{2}$, which gives $u = \frac{f}{2}.$
Substituting $f = 20\,\text{cm}$:
Case 1 (larger distance): $x_{1} = \frac{3f}{2} = \frac{3 \times 20}{2} = 30\,\text{cm}.$
Case 2 (smaller distance): $x_{2} = \frac{f}{2} = \frac{20}{2} = 10\,\text{cm}.$
Step 5: Find the ratio $x_{1} : x_{2}$
Therefore, the ratio of the two object distances is
$
\frac{x_{1}}{x_{2}}
= \frac{30}{10}
= 3:1.
$
Final Answer
The ratio of $x_{1}$ and $x_{2}$ is $3 : 1.$
