Q. An object 2.4 $m$ in front of a lens forms a sharp image on a film $12\ cm$ behind the lens. A glass plate $1cm$ thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus on film
A) $7.2m$
B) $2.4m$
C) $3.2m$
D) $5.6m$
Explanation
(c) $\frac{1}{f} = \frac{1}{12} + \frac{1}{240} = \frac{20 + 1}{240} \Rightarrow f = \frac{240}{21}m$
Shift $= 1\left( 1 - \frac{2}{3} \right) = \frac{1}{3}$
Now $v^{'} = 12 - \frac{1}{3} = \frac{35}{3}cm$
$\therefore\frac{21}{240} = \frac{3}{35} - \frac{1}{u}$
$\Rightarrow \frac{1}{u} = \frac{3}{35} - \frac{21}{240} = \frac{1}{5}\left( \frac{3}{7} - \frac{21}{48} \right)$
$\frac{5}{u} = \left| \frac{144 - 147}{48 \times 7} \right|$
$u = 560cm = 5.6m$
Hence shifting of object $= 5.6 - 2.4 = 3.2m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.