Question

An object is placed 14.0 cm in front of a lens of focal length 5.08 cm. Another lens of focal length 4.48 cm is placed 1.83 cm past the first lens. Where is the final image (how far past the lens with focal length 4.48 cm)? 2.59 cm You are correct. Your receipt no. is 158-1618 Priwinuti Tiit Is it real or virtual? Incorrect The image is virtual Correct: The image is real You are correct. Your receipt no. is 158-1963Previous Tries What is the overall magnification? 1.17 Submit Answer Incorrect. Tries 4/99 Previous Tries Screen Shot 2018-04-28 at 5.04.33 PM Search Two converging lenses are placed 81.5 cm apart. An object is placed 1.18 m to the left of the first lens, which has a focal length of 23.6 cm. The final image is located 13.2 cm to the right of the second lens. What is the focal length of the second lens? 1.05x101 m You are correct Your receipt no. is 158-236 What is the total magnification? 0.0653 Submit AnswerI Tries 5/99 Previous Tries

Answer 1

for first lens,

f1 = 5.08 cm

object distance, p1 = 14 cm

1/f = 1/p + 1/q

1/5.08 = 1/14 + 1/q1

q1 = 7.973 cm  

Now this image will work as object for second lens.

p2 = 1.83 - 7.973 = - 6.17 cm

1/4.48 = 1/(-6.16) + 1/q2

q2 = 2.6 cm ......Ans

rays actually so image is real.

m = m1 m2 = (-q1/p1)(-q2/p2)

m = (7.973/14)(2.6/-6.17)

m = - 0.24 .....Ans

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for first lens,

1/23.6 = 1/118 + 1/q1

q1 = 29.5 cm

now p2 = 81.5 - 29.5 = 52 cm

and q2 = 13.2 cm

1/f2 = 1/52 + 1/13.2

f2 = 10.5 cm .......Ans

M = m1 m2 = (-q1/p1)(-q2/p2)

M = (29.5/118)(13.2/52) = 0.0635 ......Ans