For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass plate of refractive index 1.5 and thickness 1 cm is introduced between 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?
Solution
<p>The shift produced by the glass plate is</p>
<p>$$d = t\left( {1 - {1 \over \mu }} \right) = 1 \times \left( {1 - {1 \over {1.5}}} \right) = {1 \over 3}$$ cm</p>
<p>So final image must be produced at $\left( {12 - {1 \over 3}} \right)$ cm $= {{35} \over 3}$ cm from lens so that glass plate must shift it to produce image at screen. So</p>
<p>$${1 \over {12}} - {1 \over { - 240}} = {1 \over f} = {1 \over {35/3}} - {1 \over u}$$</p>
<p>${1 \over u} = {3 \over {35}} - {1 \over {12}} - {1 \over {240}}$</p>
<p>or $u = - 560$ cm</p>
<p>so shift $= 5.6 - 2.4 = 3.2$ m</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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