Q. 114.4( 34 Votes )

# A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. 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 (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Answer :

Given:

Focal length of the objective lens, f_{1} = 2.0 cm

Focal length of the eyepiece lens, f_{2} = 6.25 cm

Distance between eyepiece and objective, d = 15 cm

We need to find the distance of object from objective lens for the following two cases:

(a) The final image is formed at least distance of distinct vision, v_{2} = 25 cm

Let the distance of object from eyepiece lens = u_{2} cm

By using lens maker formula,

…(1)

From equation 1 we have,

…(2)

Where, f_{2} = focal length of the eye piece lens

v_{2} = Distance of image formation

u_{2} = Distance of object from eyepiece

Plugging the values in equation (2)

⇒

⇒ u_{2} = -5 cm

We understand that image distance from objective lens v_{1}, is given by

v_{1} = d + u_{2}

v_{1} = 15cm - 5cm

v_{1} = 10 cm

Let image distance for the objective lens = u_{1}

From equation (1), for objective lens, we have,

…(3)

Plugging the values in equation (3) we have,

⇒ u_{1} = -2.5 cm

We know that magnifying power of a compound lens is given by,

m = ) …(4)

Where,

m = magnification

v_{1} = image distance from objective lens.

u_{1} = object distance from objective lens

d^{’} = minimum distance of distinct vision (25 cm)

f_{2} = Focal length of eyepiece

Now putting the values in equation (4)

m = 20.

Hence the magnification of the compound microscope in this case is 20. Magnification is a unit less quantity, as you can observe from the equation.

(b) Distance of final image formed, v_{2} = ∞

Object distance for the eyepiece = u_{2}

By using lens maker formula we have,

…(1)

From equation 1 we have,

…(2)

Where, f_{2} = focal length of the eye piece lens

v_{2} = Distance of image formation

u_{2} = Distance of object from eyepiece

Plugging the values in equation (2)

We get,

u_{2} = -6.25 cm

We understand that image distance from objective lens v_{1}, is given by

v_{1} = d + u_{2}

v_{1} = 15cm – 6.25cm

v_{1} = 8.75 cm

Let image distance for the objective lens = u_{1}

From equation (1), for objective lens, we have,

…(3)

Plugging the values in equation (3) we have,

⇒

⇒ u_{1} = -2.59 cm

We know that magnifying power of a compound lens is given by,

m = …(4)

Where,

m = magnification

v_{1} = image distance from objective lens.

u_{1} = object distance from objective lens

d^{’} = minimum distance of distinct vision (25 cm)

f_{2} = Focal length of eyepiece

Now putting the values in equation (4)

m = 13.51

Hence, the magnifying power of the microscope in this case is 13.51.

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