The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscope is to provide an angular magnification of -46 when used by a person with a normal near point (25 cm from the eye), what must be the focal length of the eyepiece

The focal length of eye piece is 6.52 cm.

Explanation:

Given that,

Angular Magnification of the microscope M = -46

the distance between the lens in microscope L= 16 cm

The focal length of objective f₀ = 1.5 cm

Normal near point N = 25 cm

Have to find focal length of eye piece f ₙ =?

The angular magnification is given by

M ≈ - (L-fₙ)N/f₀fₙ

Rearranging for fₙ

fₙ =L(1 - Mf₀/N)⁺¹

   =18/2.76

fₙ =  6.52 cm

The focal length of eye piece is 6.52 cm.

a lot of it means it can be "burned" for energy and sold at one price. if it is scarce, the price will go up