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

Solution

Here,
Distance of object from the objective lens, u₀ = –0.9 cm
Focal length of objective lens, f₀ = 0.8 cm

As,

\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}

∴

\frac{1}{v_{0}} = \frac{1}{f_{0}} + \frac{1}{u_{0}} = \frac{1}{0.8} - \frac{1}{0.9} = \frac{1}{7.2}

i.e.,

v_{0} = 7.2 cm

which is the distnce of the imge from the objective.

Now for the eyepiece, we have
Focal length of eyepiece, f_e = 2.5 cm
Image distance from eyepiece, v_e = – D = -25 cm
Object distance from eyepiece, u_e =?
Since,

\frac{1}{u_{e}} = \frac{1}{v_{e}} - \frac{1}{f_{e}} = - \frac{1}{25} - \frac{1}{2.5} = - \frac{11}{25}

i.e.,

u_{e} = - \frac{25}{11} = - 2.27 cm

Separation between the two lenses,

= v_{0} + |u_{e}| = 7.2 + 2.27 = 9.47 cm

Magnifying power,

M = M_{0} \times M_{e}

M = \frac{v_{0}}{u_{0}} (1 + \frac{D}{f_{e}}) = \frac{7.2}{0.9} (1 + \frac{25}{25})

= 8 \times 11 = 88

For Daily Free Study Material Join wiredfaculty

Wave Optics

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction