Optics Ques 78
- A convex lens of focal length $20 cm$ produces images of the same magnification 2 when an object is kept at two distances $x _1$ and $x _2\left(x _1>x _2\right)$ from the lens. The ratio of $x _1$ and $x _2$ is
(a) $5: 3$
(b) $2: 1$
(c) $4: 3$
(d) $3: 1$
(2019 Main, 9 April II)
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Answer:
Correct Answer: 78.(d)
Solution:
- Since, image formed by a convex lens can be real or virtual in nature.
Thus, $\quad m=+2$ or -2 .
First let us take image to be real in nature, then $m=-2=\frac{v}{u}$, where $v$ is image distance and $u$ is object distance. $\Rightarrow$
$$ v=-2 x _1 $$
[Taking $u=x _1$ ]
Now, by using lens equation,
$$ \begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ & \Rightarrow \quad \frac{1}{-2 x _1}-\frac{1}{x _1}=\frac{1}{20} \Rightarrow \frac{-3}{2 x _1}=\frac{1}{20} \\ & \Rightarrow \quad x _1=-30 cm \end{aligned} $$
Now, let us take image to be virtual in nature, then
$$ m=2=\frac{v}{u} \Rightarrow v=2 x _2 \quad\left[\text { Taking } u=x _2\right] $$
Again, by using lens equation,
$$ \begin{aligned} \Rightarrow \quad \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \Rightarrow \frac{1}{2 x _2}-\frac{1}{x _2}=\frac{1}{20} \\ \frac{-1}{2 x _2} & =\frac{1}{20} \Rightarrow x _2=-10 cm \end{aligned} $$
So, the ratio of $x _1$ and $x _2$ is
$$ \frac{x _1}{x _2}=\frac{-30}{-10}=3: 1 $$
Alternate Solution
Magnification for a lens can also be written as
$$ m=\frac{f}{f+u} $$
When $m=-2$ (for real image)
$$ \begin{aligned} -2 & =\frac{f}{f+x _1} \\ \Rightarrow \quad-2 f+(-2) x _1 & =f \text { or } x _1=-\frac{3 f}{2} \end{aligned} $$
Similarly, when $m=+2$ (for virtual image)
$$ +2=\frac{f}{f+x _2} \Rightarrow 2 f+2 x _2=f \text { or } x _2=\frac{-f}{2} $$
Now, the ratio of $x _1$ and $x _2$ is
$$ \frac{x _1}{x _2}=\frac{\frac{-3 f}{2}}{-f / 2}=\frac{3}{1} $$
BETA
