The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye.

Answer

B, C

Explanation

Here, `f_0 =1.0cm, f_e =6 cm,`
`V_c = 24 cm,`
Sepration = 9.8 cm to 11.8 cm.
`(1)/(upsilon_e) - (1)/(u_e) = (1)/(f_e)`
`=(1)/(-024) - (1)/(u_e) = (1)/(6)`
`rArr -(1)/(u_e) = (1)/(6) + (1)/(24) = (4+1)/(24) = (5)/(24)`
`:. U_e = -(24)/(5) cm = -4.8 cm.`
`:. upsilon_0 = 9.8 - 4.8 = 5 cm`
`:' (1)/(upsilon_0) - (1)/(u_0) = (1)/(f_0)`
`(1)/(5) - (1)/(u_0) = (1)/(1)`
`rArr = - (1)/(u_0) = 1 - (1)/(5) = (5-1)/(5)`
` rArr u_0 = -(5)/(4) cm = -1.25 cm.`
So, Magnifying power` =(upsilon_0)/(u_0) (1+ (D)/(f_e))`
`=(5)/(1.25) (1 + (24)/(6)) = 20`
Similarly, when, I = 11.8 cm, m =30
So, required range is 20 to 30.