Please explain the equations and method. Thank you!!! An object 2.00 cm high is

Question

Please explain the equations and method. Thank you!!!

An object 2.00 cm high is placed 35.4 cm to the left of a converging lens having a focal length of 30.4 cm. A diverging lens having a focal length -20.0 cm is placed 110 cm to the right of the converging lens. (Use the correct sign conventions for the following answers.) (a) Determine the final position and magnification of the final image. (Give the final position as the image distance from the second lens.) final position cm magnification (b) Is the image upright or inverted? (c) Repeat parts (a) and (b) for the case where the second lens is a converging lens having a focal length of +20.0 am. image position magnification image orientation upright inverted

Explanation / Answer

hi = 2 cm , u1 = 35.4 cm , f1 = 30.4 cm

from lens formula

1/f1 = 1/u1 +1/v1

1/30.4 = 1/35.4 +1/v1

v1 = 215.2 cm

d=110 cm

u2 = d- v1 = 110 -215.2 = -105.2 cm

f2 = -20 cm

-1/20 = -1/105.2 +1/v2

v2 = -24.7 cm

image is placed 24.7 cm infront of second lens

magnification M = m1*m2

M = (-v1/u1)(-v2/u2)

M = (215.2/35.4)(24.7/105.2)

M = 1.43

(b) M is positive so image is upright

(c)f1 = 20 cm

hi = 2 cm , u1 = 35.4 cm , f1 = 20 cm

from lens formula

1/f1 = 1/u1 +1/v1

1/20 = 1/35.4 +1/v1

v1 = 46 cm

d=110 cm

u2 = d- v1 = 110 -46 = 64 cm

f2 = -20 cm

-1/20 = 1/64 +1/v2

v2 = -15.23 cm

image is placed 15.23 cm infront of second lens

magnification M = m1*m2

M = (-v1/u1)(-v2/u2)

M = (-46/35.4)(64/15.23)

M = -54.6

(d) M is negative so image is inverted

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