Question

A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope.

OR

A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope.

If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 × 10⁶ m and the radius of the lunar orbit is 3.8 × 10⁸ m.

Solution

Focal length of the objective lens, f_o = 4 cm

Focal length of the eyepiece, f_e = 10 cm

Object distance, u_o = -6 cm

Magnifying power of a microscope is given by,

Now, using the lens formula, we have
$1 over straight f subscript straight o space equals space 1 over v subscript o space minus space 1 over u subscript o$

Negative sign shows that the image is inverted.

Length of the microscope is given by,

For eyepiece of the microscope,

We have, v_e = D = -25 cm

Here, L is the length of the microscope.

OR

Angular magnification is given by,
$Error converting from MathML to accessible text.$

where, fo is the focal length of the objective lens, and

f_e is the focal length of the eye-piece.

Given, diameter of the moon = 3.48 × 10⁶ m

Radius of the lunar orbit = 3.8 × 10⁸ m

Diameter of the image of moon formed by the objective lens is given by, d = $αf subscript straight o$
$Error converting from MathML to accessible text.$

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