The eyepiece of a compound microscope has a focal length of 2.70 cm , and the objective lens has f = 0.750 cm . if an object is placed 0.810 cm from the objective lens, calculate the distance between the lenses when the microscope is adjusted for a relaxed eye.
calculate the total magnification.

Explanation:

Image formation by objective lens ,

f = focal length of objective = .75 cm, ( positive )

object distance u = .81 cm ( negative )

v = image distance  

1/v - 1/u = 1/f

1/v + 1/.81 = 1 / .75

1/v = 1/ .75 - 1/.81 = .098765

v = 10.125.cm

Image formation by eye piece,

v = infinity ( for relaxed eye )

f ( eye piece ) = 2.70cm

u = ?

1/v - 1/u = 1/f

0 -1/u = 1/2.7

u = 2.7 cm

Total length  between lenses

= 2.7 + 10.125

= 12.825 cm

Total magnification = m₁ x m₂

m₁ is magnification by objective and m₂ is magnification by eye piece

m₁ = v/u = 10.125 / .810 = 12.5

m₂ =  1 + D / f = 1 + 25 / 2.7 = 10.25

Total magnification

= 12.5 x 10.25 = 128.125

explanation:

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we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .