JEE MAIN - Physics (2022 - 25th July Evening Shift - No. 7)
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?
0.8 m
3.2 m
1.2 m
5.6 m
Explanation
The shift produced by the glass plate is
$$d = t\left( {1 - {1 \over \mu }} \right) = 1 \times \left( {1 - {1 \over {1.5}}} \right) = {1 \over 3}$$ cm
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
$${1 \over {12}} - {1 \over { - 240}} = {1 \over f} = {1 \over {35/3}} - {1 \over u}$$
$${1 \over u} = {3 \over {35}} - {1 \over {12}} - {1 \over {240}}$$
or $$u = - 560$$ cm
so shift $$ = 5.6 - 2.4 = 3.2$$ m
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