Optics 3 Question 3

3. 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: 3. (d)

Solution:

  1. 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

$$ \frac{x _1}{x _2}=\frac{\frac{-3 f}{2}}{-f / 2}=\frac{3}{1} $$



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