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  • 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 fa

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 be shifted to be in sharp focus on film?

Option: 1

7.2 m


Option: 2

2.4 m


Option: 3

3.2 m


Option: 4

5.6 m


Answers (1)

best_answer

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

Here \: \: u= -2.4\: \: m = -2.4m= -240cm,\nu = 12cm

\therefore \: \: \frac{1}{2}= \frac{1}{12}-\frac{1}{\left ( -240 \right )}= \frac{1}{12}+\frac{1}{240}

f= \frac{240}{21}cm

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

shift = t\left ( 1-\frac{1}{\mu } \right )= 1\left ( 1-\frac{1}{1.5} \right )= 1\left ( 1-\frac{2}{3} \right )= \frac{1}{3}cm

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

{\nu }'= 12-\frac{1}{3}= \frac{35}{3}cm

Again\: using \: lens \: f\! ormula

\therefore \: \: \frac{21}{240}= \frac{3}{35}-\frac{1}{{u}'}\: \: or\:\: \frac{1}{{u}'}= \frac{3}{35}-\frac{21}{240}= \frac{1}{5}\left [ \frac{3}{7}-\frac{21}{48} \right ]

\frac{1}{{u}'}= -\frac{3}{1680}

{u}'= -560cm= -5.6m

\left | {u}' \right |= 5.6m

Posted by

chirag

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