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  • A convex lens focuses a distant object on a screen placed 10 cm away from it. A glass plate <img alt="\mathrm{(n=1.5)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28n%3D1.5%2

A convex lens focuses a distant object on a screen placed 10 cm away from it. A glass plate \mathrm{(n=1.5)} of thickness 1.5 is inserted between the lens and the screen. Where should the object be placed so that its image is again focused on the screen?

Option: 1

150 \mathrm{~cm}


Option: 2

160 \mathrm{~cm}


Option: 3

180 \mathrm{~cm}


Option: 4

190 \mathrm{~cm}


Answers (1)

best_answer

The situation when the glass plate is inserted between the lens and the screen, is shown in figure. The lens forms the image of object \mathrm{O} at point \mathrm{I_{1}} but the glass plate intercepts the rays and forms the final image at I on the screen. The shift in the position of image after insertion of glass plate

\mathrm{I_{1}I= t\left ( 1-\frac{1}{n} \right )= \left ( 1.5 cm \right )\left ( 1-\frac{1}{1.5} \right )= 0.5cm.}

Thus, the lens forms the image at a distance of \mathrm{9.5 \mathrm{~cm}} from itself. Using
\mathrm{\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}}, we get
\mathrm{\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{9.5}-\frac{1}{10} \quad \, or \: u=+190 \mathrm{~cm}}.

i.e. the object should be virtual and to be taken at a distance of 190 cm. from the lens right of it.

Posted by

Sayak

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