Question

An object $2.4\text{m}$ in front of a lens forms a sharp image on a film $12\text{cm}$ behind the lens. A glass plate $1\text{cm}$ thick, of refractive index $1.5$ is inserted between lens and film with its plane faces parallel to film. At what distance (from lens) should the object shifted to be in sharp focus of film?

• A. $7.2\text{m}$
• B. $2.4\text{m}$
• C. $3.2\text{m}$
• D. $5.6\text{m}$

Answer 1

The correct option is D $5.6\text{m}$


According to thin lens formula
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

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

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

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

When a glass plate is inserted between lens and film, shift produced by it will be
Shift $S=t(1-\frac{1}{\mu })=1(1-\frac{1}{1.5})=1(1-\frac{2}{3})=\frac{1}{3}\text{cm}$

To get image at film, lens should form image at distance

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

Again using lens formula,

$\therefore \frac{21}{240}=\frac{3}{35}-\frac{1}{u^{\prime }}$

$\Rightarrow \frac{1}{u^{\prime }}=\frac{3}{35}-\frac{21}{240}=\frac{1}{5}[\frac{3}{7}-\frac{21}{48}]$

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

$u^{\prime }=-560\text{cm}\text{or}-5.6\text{m}$

$\therefore |u^{\prime }|=5.6\text{m}$

Hence, option (d) is the correct answer.