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For an object placed at a distance $2.4 \mathrm{~m}$ from a lens, a sharp focused image is observed on a screen placed at a distance $12 \mathrm{~cm}$ form the lens. A glass plate of refractive index $1.5$ and thickness $1 \mathrm{~cm}$ is introduced between the lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?
Option: 1
$0.8\: \mathrm{m}$

Option: 2
$3.2\: \mathrm{m}$

Option: 3
$1.2\: \mathrm{m}$

Option: 4
$5.6\: \mathrm{m}$

Answers (1)

$\mathrm{u=-2.4\mathrm{~m}=-240 \mathrm{~cm} }$

$\mathrm{v=+12 \mathrm{~cm} }$

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

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

$\mathrm{-\frac{240}{21}=\frac{80}{7} \mathrm{~cm}}$

Normal shift due to glass plate

$\mathrm{s =t\left(1-\frac{1}{\mu}\right) }$

$\mathrm{= 1 \mathrm{~cm}\left(1-\frac{1}{3 / 2}\right) }$

$\mathrm{s=\frac{1}{3} \mathrm{~cm} }$

$\mathrm{v^{\prime} =v-s=12-\frac{1}{3}=\frac{35}{3} \mathrm{~cm} }$

$\mathrm{\frac{1}{v^{\prime}}-\frac{1}{u^{\prime}} =\frac{1}{f} }$

$\mathrm{\frac{1}{+\left(\frac{35}{3}\right)}-\frac{1}{u^{\prime}} =\frac{1}{\frac{80}{7}} }$

$\mathrm{\frac{1}{u^{\prime}} = \frac{3}{35}-\frac{7}{80}}$

$\mathrm{=\frac{24}{280}-\frac{24.5}{280}}$

$\mathrm{\frac{1}{u'}=\frac{-1}{560}}$

$\mathrm{|u'|=560\: cm=5.6 \ m}$

The object is displaced by=5.6 - 2.4

$\mathrm{=3.2\: m}$

Hence (2) is correct option

Posted by

Ramraj Saini

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