The first lens
do = 121, f1 = 25.9
1/di + 1/do =1/f1,
di = 32.9 cm
It is the image for the second lens. It is 41.2 – 32.9 = 8.25 cm far from the second lens.
do=8.25, f2 =15.8
1/do -1/di =-1/f2.
di =13.2 cm.
Again this would work as image for the 3rd lens.
It is 51.2 -13.2 = 28cm from the origin of the coordinate system and distance 51.4 – 28 = 23.4 cm from the 3rd lens.
do =23.4, f3 = 10
1/di + 1/do = 1/f3
di =17.5.
The final image is 17.5 + 51.4 =68.9 cm from the origin of the coordinate system.
M = di/do, and M =M1•M2 •M3=
=(32.9/121) •(13.2/8.25) •(17.5/23.4) =0.325.
Physics lenses. Please help, my math is wrong and I need second opinion.?
Consider three lenses with focal lengths of 25.9 cm, -15.8 cm, and 10.0 cm positioned on the x axis at x1 = 0 m, x2 = 0.412 m, and x3 = 0.514 m, respectively. An object is at d = -121 cm.
Find the location of the final image produced by this lens system. x=___m
Find the magnification of the final image produced by this lens system. m =____
work is here
imgur. com/ned2q
3 answers
If you know the answer, show it. please
How much work is required to lift a 5.1-kg concrete block to a height of 3.6 m
2 answers