An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

Solution

Given, angular magnification = 30
Focal length of objective, f_o = 1.25 cm
Focal length of eyepiece, f_e = 5 cm

In normal adjustment, image is formed at least distance of distinct vision, d = 25 cm.

Using formula,
Angular magnification of eyepiece =

(1 + \frac{D}{f_{e}})

= (1 + \frac{25}{5}) = 6

Now, magnification of objective lens,

m = \frac{30}{6} = 5

∴

m = - \frac{v_{0}}{u_{0}} = 5 or, v_{0} = - 5 u_{0}

For objective lens, using the lens formula,

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

∴

\frac{1}{- 5 μ_{0}} - \frac{1}{μ_{0}} = \frac{1}{1.25}

\frac{- 6}{5 μ_{0}} = \frac{1}{1.25}

u_{0} = - \frac{6 \times 1.25}{5} = - 1.5 cm

i.e., object should be held at a distance 1.5 cm in front of objective lens.

As,

v_{0} = - 5 μ_{0}

∴

v_{0} = - 5 (- 1.5) = 7.5 cm

For eyepiece lens,

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

[using len's formula]
we have,

\frac{1}{μ_{e}} = \frac{1}{v_{e}} - \frac{1}{f_{e}} = \frac{1}{- 25} - \frac{1}{5} = - \frac{6}{25}

u_{e} = - \frac{25}{6} = - 4.17 cm

Thus, the object is at a distance of 4.17 cm from the eyepiece.

Separation between the objective lens and eyepiece = | u_e| + | v₀ | = 4.17 + 7.5 = 11.67 cm.

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Wave Optics

An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

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