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QJEEM12Phe7b9: JEE Main 2012 Physics - Class 12 JEEM12Phe7b9 Solution: An object in front of a lens forms a sharp image on a film…

An object 2.42.4 mm in front of a lens forms a sharp image on a film 1212 cmcm behind the lens. A glass plate 11 cmcm thick, of refractive index 1.501.50 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?

Multiple Choice Options

  • A 3.23.2 mm
  • B 7.27.2 mm
  • C 2424 mm
  • D 5.65.6 mm

Step-by-Step Verified Solution

The focal length of the lens

1f=1υ−1u=112+1240{1 \over f} = {1 \over \upsilon } - {1 \over u} = {1 \over {12}} + {1 \over {240}}

=20+1240=21240= {{20 + 1} \over {240}} = {{21} \over {240}}

f=24021cmf = {{240} \over {21}}cm

Shift =t(1−1μ)⇒1(1−13/2)= t\left( {1 - {1 \over \mu }} \right) \Rightarrow 1\left( {1 - {1 \over {3/2}}} \right)

=1×13= 1 \times {1 \over 3}

Now v′=12−13=353cmv' = 12 - {1 \over 3} = {{35} \over 3}cm

Now the object distancce u.u.

1u=335−21240=15[37−2148]{1 \over u} = {3 \over {35}} - {{21} \over {240}} = {1 \over 5}\left[ {{3 \over 7} - {{21} \over {48}}} \right]

1u=15[48−497×16]{1 \over u} = {1 \over 5}\left[ {{{48 - 49} \over {7 \times 16}}} \right]

u=−7×16×5=−560cm=−5.6 mu = - 7 \times 16 \times 5 = - 560cm = - 5.6\,m

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