Question

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

Answer 1

Answer:

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

= $\frac{v_o}{u_o} \times\frac{D}{f_e}$

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