A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Focal length of the objective lens, f1 = 2.0 cm
Focal length of the eyepiece, f2 = 6.25 cm
Distance between the objective lens and the eyepiece, d = 15 cm
(a) Least distance of distinct vision, $d^{\prime}=25 \mathrm{~cm}$
∴ Image distance for the eyepiece, v2 = −25 cm
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$
$\frac{1}{u_{2}}=\frac{1}{v_{2}}-\frac{1}{f_{2}}$
$=\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$
$\therefore u_{2}=-5 \mathrm{~cm}$
Image distance for the objective lens, $v_{1}=d+u_{2}=15-5=10 \mathrm{~cm}$
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
$\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$
$=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$
$\therefore u_{1}=-2.5 \mathrm{~cm}$
Magnitude of the object distance, $\left|u_{1}\right|=2.5 \mathrm{~cm}$
The magnifying power of a compound microscope is given by the relation:
$m=\frac{v_{1}}{\left|u_{1}\right|}\left(1+\frac{d^{\prime}}{f_{2}}\right)$
$=\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=4(1+4)=20$
Hence, the magnifying power of the microscope is 20.
(b) The final image is formed at infinity.
$\therefore$ Image distance for the eyepiece, $v_{2}=\infty$
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$
$\frac{1}{\infty}-\frac{1}{u_{2}}=\frac{1}{6.25}$
$\therefore u_{2}=-6.25 \mathrm{~cm}$
Image distance for the objective lens, $v_{1}=d+u_{2}=15-6.25=8.75 \mathrm{~cm}$
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
$\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$
$=\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$
$\therefore u_{1}=-\frac{17.5}{6.75}=-2.59 \mathrm{~cm}$
Magnitude of the object distance, $\left|u_{1}\right|=2.59 \mathrm{~cm}$
The magnifying power of a compound microscope is given by the relation:
$m=\frac{v_{1}}{\left|u_{1}\right|}\left(\frac{d^{\prime}}{\left|u_{2}\right|}\right)$
$=\frac{8.75}{2.59} \times \frac{25}{6.25}=13.51$
Hence, the magnifying power of the microscope is 13.51.
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- JEE Main
- Exam Pattern
- Previous Year Papers
- PYQ Chapterwise
- Physics
- Kinematics 1D
- Kinemetics 2D
- Friction
- Work, Power, Energy
- Centre of Mass and Collision
- Rotational Dynamics
- Gravitation
- Calorimetry
- Elasticity
- Thermal Expansion
- Heat Transfer
- Kinetic Theory of Gases
- Thermodynamics
- Simple Harmonic Motion
- Wave on String
- Sound waves
- Fluid Mechanics
- Electrostatics
- Current Electricity
- Capacitor
- Magnetism and Matter
- Electromagnetic Induction
- Atomic Structure
- Dual Nature of Matter
- Nuclear Physics
- Radioactivity
- Semiconductors
- Communication System
- Error in Measurement & instruments
- Alternating Current
- Electromagnetic Waves
- Wave Optics
- X-Rays
- All Subjects
- Physics
- Motion in a Plane
- Law of Motion
- Work, Energy and Power
- Systems of Particles and Rotational Motion
- Gravitation
- Mechanical Properties of Solids
- Mechanical Properties of Fluids
- Thermal Properties of matter
- Thermodynamics
- Kinetic Theory
- Oscillations
- Waves
- Electric Charge and Fields
- Electrostatic Potential and Capacitance
- Current Electricity
- Thermoelectric Effects of Electric Current
- Heating Effects of Electric Current
- Moving Charges and Magnetism
- Magnetism and Matter
- Electromagnetic Induction
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- Ray Optics and Optical Instruments
- Wave Optics
- Dual Nature of Radiation and Matter
- Atoms
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