Home
Class 12
PHYSICS
A compound micrscope consists of an obje...

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.

Text Solution

Verified by Experts

a. For the eyepiece `v_e=-25cm nd f_e=+5cm`
Using `1/v_e-1/u_e=1/f_e`
`1/u_e=1/v_e-1/f_e`
`-1/(25cm_-1/(5cm)`
or `u_e=25/6cm=-4.17=-4.2cm`
As the objective is 12.2 cm away from the eyepiece the image formed by the objective is 12.2 cm-4.2 cm =8.0 cm away from it. for the objective
`v=+8.0 cm, f_0=+1.0cm`
Using `1/v-1/u=1/f_0`
`1/u=1/v-1/f_0`
`=1/(8.0cm)-1/(1.0cm)`
or `u=-8.0/7.0cm=-1.1cm.`
b. The angular magnification is
`m=v/u(1+D/f_e)`
`=(+8.0cm)/(-1.1cm)(1+(25cm)/(5cm))=-44`.
Promotional Banner

Topper's Solved these Questions

  • OPTICAL INSTRUMENTS

    HC VERMA|Exercise Short Answer|11 Videos
  • OPTICAL INSTRUMENTS

    HC VERMA|Exercise Objective 1|9 Videos
  • OPTICAL INSTRUMENTS

    HC VERMA|Exercise Exercises|23 Videos
  • MAGNETIC PROPERTIES OF MATTER

    HC VERMA|Exercise Exercises|9 Videos
  • PERMANENT MAGNETS

    HC VERMA|Exercise Exercises|25 Videos