In a compound microscope, the objective and eyepiece have focal lengths 0.95 cm and 5 cm respectively and are kept at a distance of 20 cm .the final image is formed at a distance of 25 cm from the eyepiece. Calculate the position of the object and the total magnification .

Answer

D

Explanation

For the given compound microscope, `f_0 = (1)/(20) D = 0.05m - 5 cm,`
`f_e = (1)/(10)D = 0.1 m =10 cm`
D = 25 cm, seperation between objective and eyepiece = 20 cm For the minimum seperation between two points which can be distinguished by eye using the miscroscope, the magnifiying power should maximum. For the eyepiece.
`upsilon_e = -25 cm f_e = 10 cm`
So, ` (1)/(u_e) = (1)/(upsilon_e) - (1)/(f_e)`
` =(1)/(-25) - (1)/(10) = -(2+5)/(50)`
`rArr u_e = (-50)/(7) cm`
So the image distance for the objective lens should be,
`upsilon_0 = 20 - (50)/(7) = (90)/(7)cm`
Now, for the objective lens,
`(1)/(u_0) = (1)/(upsilon_0) - (1)/(f_0)`
`=(7)/(90) - (1)/(5) = (7-18)/(90) =- (11)/(90)`
`rArr u_0 = (-90)/(11)cm`
`m =(upsilon_0)/(u_0) (1+ (D)/(f_e))`
`=((90)/(7))/((-90)/(11)) (1+ (25)/(10))`
`=(11)/(7)xx (3.5) = 5.5`
Thus, minimum separation that the eye can distinguish `= (0.22)/(5.5) mm = 0.4 mm.`