JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 22)
An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification '6', gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to __________ cm.
Answer
25
Explanation
Given, magnification, M = 6
Since we know that magnifying power of a simple microscope is given by
$$M = 1 + {D \over {{f_0}}}$$
where, D = least distance of distinct vision = 25 cm
and f0 = focal length of objective lens.
$$ \Rightarrow 6 = 1 + {D \over {{f_0}}} \Rightarrow 6 = 1 + {{25} \over {{f_0}}} \Rightarrow 5 = {{25} \over {{f_0}}} \Rightarrow {f_0} = 5$$ cm
For compound microscope, magnifying power is given by
$$M = {{I\,.\,D} \over {{f_0}{f_e}}} = 2$$Msimple microscope
where, f0 = fe are the focal lengths of the objective lens and eye piece respectively
and l = length of the given tube = 0.6 m
$$ \Rightarrow 12 = {{60 \times 25} \over {5\,.\,{f_e}}}$$ [$$\because$$ magnification is doubled]
$$\Rightarrow$$ fe = 25 cm
This is the required focal length of eyepiece.
Since we know that magnifying power of a simple microscope is given by
$$M = 1 + {D \over {{f_0}}}$$
where, D = least distance of distinct vision = 25 cm
and f0 = focal length of objective lens.
$$ \Rightarrow 6 = 1 + {D \over {{f_0}}} \Rightarrow 6 = 1 + {{25} \over {{f_0}}} \Rightarrow 5 = {{25} \over {{f_0}}} \Rightarrow {f_0} = 5$$ cm
For compound microscope, magnifying power is given by
$$M = {{I\,.\,D} \over {{f_0}{f_e}}} = 2$$Msimple microscope
where, f0 = fe are the focal lengths of the objective lens and eye piece respectively
and l = length of the given tube = 0.6 m
$$ \Rightarrow 12 = {{60 \times 25} \over {5\,.\,{f_e}}}$$ [$$\because$$ magnification is doubled]
$$\Rightarrow$$ fe = 25 cm
This is the required focal length of eyepiece.
Comments (0)
