Question

A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm, and that of the eyepiece is 1.0 cm. The near-point distance of the person using the microscope is 25.0 cm. The magnitude of the final magnification of the microscope is closes to:

Answer 1

m = 14*26 = 364

Step-by-step explanation:

overall magnification is given as m

`m = m_(o)* m_(e)`

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

`m_(o) = (v_(o))/(-u _(0))`

and me as

`m_(e) = 1+(D)/(f_(e))`

d is distant of distinct vision = 25.0 cm for normal eye

fe = focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

`(1)/(v_(0))-(1)/(u_(0))=(1)/(f_(0))`

`(1)/(v_(0)) = (1)/(u_(0)) + (1)/(f_(0))`

`(1)/(v_(0)) = (1)/(0.150) + (1)/(0.14)`

`(1)/(v_(o)) = 2.1 cm`

`m_(o) = (2.1)/(0.150) = 14`

`m_(e) = 1+(25)/(1)`

`m_(e) =26`

`m = m_(o)* m_(e)`

m = 14*26 = 364