An object $2.4 \mathrm{m}$ in front of a lens forms a sharp image on a screen $12 \mathrm{cm}$ behind the lens. A glass plate $1 \mathrm{cm}$ thick, of refractive index $1.50$ is interposed between lens and screen with its plane faces parallel to screen. At what distance from the lens, should object be placed so as to get a sharp image on the screen?

• A

  $2.4 \mathrm{m}$

   
• B

  $3.2 \mathrm{m}$

   
• Correct answer
  C

  $5.6 \mathrm{m}$

   
• D

  $7.2 \mathrm{m}$

   

Solution

According to the thin lens formula

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

Here, $\mathrm{u} = -2.4 \mathrm{m} = - 240 \mathrm{cm}, \mathrm{v} = 12 \mathrm{cm}$

$\therefore$  $\frac{1}{f}=\frac{1}{12}-\frac{1}{\left(-240\right)}=\frac{1}{12}+\frac{1}{240}$

       $\frac{1}{f}=\frac{21}{240}\text{ or }f=\frac{240}{21} \text{cm}$

When a glass plate is interposed between lens and film, so shift produced by it will be

Shift $=\text{t }\left(1-\frac{1}{\mu }\right)=1\left(1-\frac{1}{1\text{.}5}\right)=1\left(1-\frac{2}{3}\right)=\frac{1}{3} \text{cm}$

To get an image at the film, the lens should form an image at a distance

              $v{}^{\text{'}}=12-\frac{1}{3}=\frac{35}{3} \text{cm}$

Again using lens formula

$\therefore$  $\frac{21}{240}=\frac{3}{35}-\frac{1}{u^{\prime }}\text{ or }\frac{1}{u^{\prime }}=\frac{3}{35}-\frac{21}{240}=\frac{1}{5}\left[\frac{3}{7}-\frac{21}{48}\right]$

      $\frac{1}{u^{\prime }}=\frac{1}{5}\left[\frac{144-147}{336}\right]\text{ or }\frac{1}{u^{\prime }}=-\frac{3}{1680}$

     $u{}^{\text{'}}=-56\mathrm{0 }\text{cm}=-5\text{.}6 \text{m}$

     $\left|u{}^{\text{'}}\right|=5\text{.}6 \text{m}$

Hence, the correct answer is option: 'C'.