The focal length of the objective and eye lens of a compound are 2 cm and 6.25 cm respectively. The distance between the lenses is 15 cm. How far the objective lens will be object be kept so as to obtain the final image at the near point of the eye. Also calculate its magnifying power.

200 cm

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160 cm

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2.5 cm

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0.1 cm

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Solution

The correct option is C 2.5 cm
length of the objective lens $(f_{1}) = 2.0 c m$
Focal length of the eyepiece $(f_{2}) = 6.25 c m$
Distance between the objective lens and the eyepiece $(d) = 15 c m$
$(a)$ Least distance of distinct vision $, (d_{1}) = 25 c m$
Therefore, Image distance for the eyepiece $(v_{2}) = - 25 c m$
Object distance for the eyepiece $= u_{2}$
According to the lens formula $,$
$1 / v_{2} - 1 / u_{2} = 1 / f_{2}$
$1 / u_{2} = 1 / v_{2} - 1 / f_{2} = 1 / - 25 - 1 / 6.25 = - 1 - 4 / 25 = - 5 / 25$
$u_{2} = - 5 c m$
Image distance for the objective lens $, (v_{1}) = - d + u_{2} = 15 - 5 = 10 c m$
Object distance for the objective lens $= u_{2}$
According to the lens formula $,$
$1 / v_{1} - 1 / u_{1} = 1 / f_{1}$
$1 / u_{1} = 1 / v_{1} - 1 / f_{1}$
$= 1 / 10 - 1 / 2 = 1 - 5 / 10 = - 4 / 10$
$u_{1} = - 2.5 c m$
Magnitude of the object distance $(u_{1}) = 2.5 c m$
Hence,
option $(C)$ is correct answer.

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