A compound micrscope consists of an objective of focal length 1.0 cm and an eyepiece of focal length 5.0 cm separated by 12.2 c.a. At wht distance from the objective should a object be placed to focus it prope4y so that the final image is formed at the least distance of clear vision (25cm)? b. calculate the angular magnificationn in this case.

Explanation

Image distance from the eyepiece, `v_(e)=-25cm`
Focal length of eyepiece, `f_(e)=6.25cm`
Using lens formula,
`-(1)/(25)-(1)/(u_(e))=(1)/(6.25)`
`(1)/(u_(e))=-(1)/(25)-(1)/(6.25)`
`u_(e)=-5cm`
Distance between the objective and eyepiece = 15 cm
Now, image distance of image formed by the objective,
`v_(o)=15-u_(e)=15-5=10cm`
Focal length of objective, `f_(o)=2.0 cm`
`rArr" "(1)/(v_(o))-(1)/(u_(0))=(1)/(f_(o))`
`(1)/(10)-(1)/(u_(o))=(1)/(2)`
`(1)/(u_(o))=(1)/(10)-(1)/(2)`
`=(1-5)/(10)=-(4)/(10)`
`u_(o)=(-10)/(4)=-2.5cm`
This is the distance of object from the objective
Magnifying power `=(v_(o))/(|u_(o)|)(1+(D)/(f_(e)))`
`=(10)/(2.5)xx(1+(25)/(6.25))`
`=(10xx31.25)/(2.5xx6.25)=(312.5)/(15.625)=20`
(b) for this case, `v_(e)=oo`
`-u_(e)=f_(e)=6.25cm rArr u_(e)=-6.25cm`
`therefore" "v_(o)=15-6.25=8.75 cm`
`f_(o)=2.0 cm`
`"Using, "(1)/(v_(o))-(1)/(u_(o))=(1)/(f_(o))`
`-(1)/(u_(o))=(1)/(f_(o))-(1)/(v_(o))`
`-(1)/(u_(o))=(1)/(2)-(1)/(8.75)=(6.75)/(17.5)`
`"There, "u_(o)=-(17.5)/(6.75)=-2.59cm`
Magnifying power
`M=(v_(o))/(|u_(o)|)xx(D )/(f_(e))=(8.75)/(2.59)xx(25)/(6.25)=13.51`