JEE MAIN - Physics (2012 - No. 5)
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 $$1$$ $$cm$$ 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 shifted to be in sharp focus of film?
$$7.2$$ $$m$$
$$24$$ $$m$$
$$3.2$$ $$m$$
$$5.6$$ $$m$$
Explanation
The focal length of the lens
$${1 \over f} = {1 \over \upsilon } - {1 \over u} = {1 \over {12}} + {1 \over {240}}$$
$$ = {{20 + 1} \over {240}} = {{21} \over {240}}$$
$$f = {{240} \over {21}}cm$$
Shift $$ = t\left( {1 - {1 \over \mu }} \right) \Rightarrow 1\left( {1 - {1 \over {3/2}}} \right)$$
$$ = 1 \times {1 \over 3}$$
Now $$v' = 12 - {1 \over 3} = {{35} \over 3}cm$$
Now the object distancce $$u.$$
$${1 \over u} = {3 \over {35}} - {{21} \over {240}} = {1 \over 5}\left[ {{3 \over 7} - {{21} \over {48}}} \right]$$
$${1 \over u} = {1 \over 5}\left[ {{{48 - 49} \over {7 \times 16}}} \right]$$
$$u = - 7 \times 16 \times 5 = - 560cm = - 5.6\,m$$
$${1 \over f} = {1 \over \upsilon } - {1 \over u} = {1 \over {12}} + {1 \over {240}}$$
$$ = {{20 + 1} \over {240}} = {{21} \over {240}}$$
$$f = {{240} \over {21}}cm$$
Shift $$ = t\left( {1 - {1 \over \mu }} \right) \Rightarrow 1\left( {1 - {1 \over {3/2}}} \right)$$
$$ = 1 \times {1 \over 3}$$
Now $$v' = 12 - {1 \over 3} = {{35} \over 3}cm$$
Now the object distancce $$u.$$
$${1 \over u} = {3 \over {35}} - {{21} \over {240}} = {1 \over 5}\left[ {{3 \over 7} - {{21} \over {48}}} \right]$$
$${1 \over u} = {1 \over 5}\left[ {{{48 - 49} \over {7 \times 16}}} \right]$$
$$u = - 7 \times 16 \times 5 = - 560cm = - 5.6\,m$$
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