Physics 1 Marks Question

$\frac{1}{\text{f}_2}=\Big(\frac{\mu_2}{\mu_\text{m}}-1\Big)\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\Rightarrow \frac{1}{\text{v}}=\frac{1}{(-12)}+\frac{1}{-\text{x}}$

$\Rightarrow \text{v}=-\Big(\frac{12\text{x}}{\text{x}+12}\Big)$

$\text{u}=-\Big(5+\frac{12\text{x}}{\text{x}+12}\Big)$

$=-\Big(\frac{17\text{x}+60}{\text{x}+12}\Big)$

$\text{f}=-7.5\text{cm}$

$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\Rightarrow \frac{1}{\text{v}}=\frac{1}{-7.5}+\frac{\text{x}+12}{17\text{x}+60}$

$\Rightarrow \frac{1}{\text{v}}=\frac{17\text{x}+60-7.5}{7.5(17\text{x}+60)}$

$\Rightarrow \text{v}=\frac{7.5(17\text{x}+60)}{52.5-127.5\text{x}}$

$\Rightarrow \text{v}=\frac{250(\text{x}+4)}{15\text{x}-100}$

$\Rightarrow \text{v}=\frac{50(\text{x}+4)}{(3\text{x}-20)}$

$\text{u}=-\Big[\frac{5-50(\text{x}+4)}{3\text{x}-20}\Big]$

$\text{v}=\pm\text{x}$ (since, the final image is produced on the object A"B")

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\Rightarrow \frac{1}{\text{x}}+\frac{1}{\frac{5-50(\text{x}\times4)}{3\text{x}-20}}=\frac{1}{-20}$

$\Rightarrow 25\text{x}^2-1400\text{x}-6000=0$

$\Rightarrow \text{x}^2-56\text{x}-240=0$

$\Rightarrow (\text{x}-60)(\text{x}+4)=0$

Let the object to placed at a distance x from the lens further away from the mirror.

For the concave lens (1^st refraction)

u = -x, f = -20cm

From lens formula,

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{-20}+\frac{1}{-\text{x}}$

$\Rightarrow\text{v}=-\Big(\frac{20\text{x}}{\text{x}+20}\Big)$

So, the virtual image due to fist refraction lies on the same side as that of object. (A'B')

This image becomes the object for the concave mirror.

For the mirror,

$\Rightarrow\text{u}=-\Big(5+\frac{20\text{x}}{\text{x}+20}\Big)=-\Big(\frac{25\text{x}+100}{\text{x}+20}\Big)$

$\text{f}=-10\text{cm}$

From mirror equation,

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{-10}+\frac{\text{x}+20}{25\text{x}+100}$

$\Rightarrow\text{v}=\frac{50(\text{x}+4)}{3\text{x}-20}$

So, this image is formed towards left of the mirror.

Again for second refraction in concave lens,

$\text{u}=-\Big[5-\frac{50(\text{x}+4)}{3\text{x}-20}\Big]$ (assuming that image of mirror is formed between the lens and mirro 3x -20)

$\text{v}=+\text{x}$ (Since, the final image is produced on the object A"B")

Using lens formula,

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\Rightarrow\frac{1}{\text{x}}+\frac{1}{\frac{5-50(\text{x}+4)}{3\text{x}-20}}=\frac{1}{-20}$

$\Rightarrow 25\text{x}^2-1400\text{x}-6000=0$

$\Rightarrow\text{x}^2-56\text{x} -240=0$

$\Rightarrow (\text{x}-60)(\text{x}+4)=0$

So, $\Rightarrow\text{x}=60\text{cm}$

The object should be placed at a distance 60cm from the lens further away from the mirror. So that the final image is formed on itself.