A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,

Explanation

Here, focal lenght of objective `f_0 = 0.8 cm `
Focal length of eye- piece `f_e = 2.5 cm`
For objective
objective distance
`u_0 = -0.9 cm `
From lens formula
`(1)/(f_0) = (1)/(v_0) - (1)/(u_0)`
`therefore (1)/(v_0) = (1)/(f_0) + (1)/(u_0) = (1)/(0.8) + (1)/(-0.9) = (0.9 - 0.8)/(0.72)`
`therefore (1)/(v_0) = (0.1)/(0.72)`
`therefore v_0 = 7.2 cm `
Now for eye - piece Image distance
`v_e = D = -25 cm`
From lens formula
, `1/f_e = (1)/(v_e) - (1)/(u_e)`
`therefore 1/(u_e) = (1)/(v_e) - (1)/(f_e) = (1)/(-25) - (1)/(2.5) =11`
`therefore (1)/(u_e) = (-1-10)/(25) = (-11)/(25)`
`therefore u_e = -(25)/(11) = -2.27 cm `
` rArr ` Distance between two lenses
`= v_0 + |u_e|`
`= 7.2 + 2.27`
` = 9.47 cm `
`rArr` Magnification m `= m_0 xx m_e`
`= (v_0)/(|u_0|) xx (1 + (D)/(f_e))`
`= (7.2)/(0.9) xx (1 I (25)/(2.5)) = 8 xx (1 + 10)`
`= 8 xx 11`
= 88