The two lenses in a microscope are separated by 19.5 cm, and the focal length of the eyepiece lens is 2.75 cm.

a) if the image of this object is viewed with the eyepiece adjusted for minimum eyestrain (image at the far point of the eye) for a person with normal vision, how far from the objective lens will the objective image need to be formed?
b) where is the object located, if the objective lens has a focal length of 0.350 cm?
c) what is the angular magnification of this microscope? (approximation)
d) if a near sighted person with a far point of 1.95 m looked through this microscope what would the distance between the two lenses need to be to get minimal eyestrain?
e) and what would that magnification be? (approximation)

Explanation:

a )  If the image of this object is viewed with the eyepiece adjusted for minimum eyestrain (image at the far point of the eye) , the image from object lens must have been formed at the focus of eye lens . So the objective image must have been formed at 19.5 - 2.75 = 16.75 cm from the object lens.

b ) Let the object distance be u

For object lens

v = 16.75 cm , f = .35 cm

1/v - 1/u = 1/f

1/16.75  - 1/u = 1/ .35

.0597 - 1/u = 2.857

1/u = - 2.7973

u = .3575 cm

c ) Angular magnification

=

v₀ and u₀ are image and object distance for object lens , D = 25 cm and f_e is focal length of eye lens

= (16.75 / .3575) x( 25 / 2.75)

= 46.85 x 9.09

= 426

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s