Question

An object $2.4m$ in front of a lens forms a sharp image on a screen $12cm$ behind the lens. A glass plate $1cm$ 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. $2.4m$
• B. $3.2m$
• C. $5.6m$
• D. $7.2m$

Answer 1

Answer: (C)

According to the thin lens formula
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
Here, $u=-2.4m=-240cm,v=12cm$
$\therefore$ $\frac{1}{f}=\frac{1}{12}-\frac{1}{\left(-240\right)}=\frac{1}{12}+\frac{1}{240}$
$\frac{1}{f}=\frac{21}{240}\text{ or }f=\frac{240}{21}\text{cm}$
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
$v^{{}^{\prime }}=12-\frac{1}{3}=\frac{35}{3}\text{cm}$
Again using lens formula
$\therefore$ $\frac{21}{240}=\frac{3}{35}-\frac{1}{u^{{}^{\prime }}}\text{ or }\frac{1}{u{}^{{}^{\prime }}}=\frac{3}{35}-\frac{21}{240}=\frac{1}{5}\left[\frac{3}{7}-\frac{21}{48}\right]$
$\frac{1}{u^{{}^{\prime }}}=\frac{1}{5}\left[\frac{144-147}{336}\right]\text{ or }\frac{1}{u{}^{{}^{\prime }}}=-\frac{3}{1680}$
$u^{{}^{\prime }}=-560\text{cm}=-5\text{.}6\text{m}$
$\left|u^{{}^{\prime }}\right|=5\text{.}6\text{m}$