An eye can distinguish between two points of an object if they are separated by more than 0.22 mm when the object is placed at 25 cm from the eye. The object is now seen by a compound microscope having a 20 D objective and 10 D eyepiece separated by a distance of 20 cm. The final image is formed at 25 cm from the eye. What is the minimum separation between two points of the object which can now be distinguished?

Options

A. 4000J,3A
B. 8000 J,3A
C. 4000 J,4A
D. 8000 J,5A

Answer

A

Explanation

Magnifying power of microscope
`u_(c)=(Df_(c))/(D+f_(c))=(10xx25)/(10+25)=(50)/(7)cm`
`v_(0)=20-(50)/(7)=(90)/(7)cm`
`m_("obj")=(|f-v|)/(f)-((90)/(7)-5)/(5)=(11)/(7)cm`
`m_("eye")=1+(D)/(f_(c))=1+2.5=3.5`
`M=m_("obj")xxm_("eye")=3.5xx(11)/7)=(11)/2)`
New resolving limit `=(0.22)/(11//2)mm=0.04mm`