Which could describe the motion of any object
1 answer:
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Answer:
check the image above please
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
=
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