A converging lens has a focal length of 13.5cm . For each of two objects located

Question

A converging lens has a focal length of 13.5cm . For each of two objects located to the left of the lens, one at a distance of s1 = 20.0cm and the other at a distance of s2 = 4.50cm , determine

A. the image position

S1=?

s2=?

B. the magnification

m1=?

m2=?

C. whether the image is real or virtual

a.Both images are real.

b.The image of the first object is real and the image of the second object is virtual.

c.The image of the first object is virtual and the image of the second object is real.

d.Both images are virtual.

D. whether the image is errect or inverted

a.Both images are erect.

b.The image of the first object is erect and the image of the second object is inverted.

c.The image of the first object is inverted and the image of the second object is erect.

d.Both images are inverted.

Please explain I really confused on this :/. Thank you!

Explanation / Answer

1/S + 1/S' = 1/F===>S' = (SF)/(S-F)

A)

1/20 + 1/S1' = 1/13.5

S1' = (20*13.5)/(20-13.5) = 41.5385 cm

S2' = (4.5*13.5)/(4.5-13.5) = ?6.75 cm

B) m1 = -S1'/S1 = (-41.5385/20) = ?2.0769

m2 = S2'/S2 = 6.75/4.5 = 1.5

C) b.The image of the first object is real and the image of the second object is virtual.

D) The image of the first object is inverted and the image of the second object is erect.

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