Question

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As an aid in working this problem, consult Concept Simulation 26.3. A converging lens has a...

As an aid in working this problem, consult Concept Simulation 26.3. A converging lens has a focal length of 87.00 cm. A 12.0 cm tall object is located 155.0 cm in front of this lens.

(a) What is the image distance? cm

(b) Is the image real or virtual?

real virtual?

(c) What is the image height? Be sure to include proper algebraic sign.
cm?

Expert Solution

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

As an aid in working this problem, consult Concept Simulation 26.3.

A converging lens has a focal length of 88.50 cm. A 11.0 cmtall object is located 165.0 cm infront of this lens.

(a) What is the image distance?

(b) Is the image real or virtual?real virtual

(c) What is the image height? Be sure to include proper algebraicsign.

focal length of the lens f = 88.5 cm

height of the object h _o = 11 cm

object distance d_o = 165 cm

Thin lens equation is

1 / d_o + 1 / d _i = 1 /f

1 / 165 cm + 1 / d _i = 1 / 88.5 cm

1 / d _i = 1 /88.5 - 1 / 165 cm

image distance d _i = 190.88 cm = 1.90 m

The image is real image

magnification of the image M = - d_i / d_o

= -190.88 / 165 = 1.156

height of the image h_i = M h _o = 1.156 * 11 = 12.7253 cm

genius_generous answered 4 months ago

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