MasteringPhysics: Assignment 30 Google Chrome Secure https://session.masteringph

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MasteringPhysics: Assignment 30 Google Chrome Secure https://session.masteringphysics.com/myct/itemView?assignmentProblemID-791349828offset-prev Assignment 30 Problem 35.30 a previous | 3 of 4 | next» Problem 35.30 Part C What is the distance between the final image and the lens? Express your answer to two significant figures and include the appropriate units. Value cm Figure 1 Submit My Answers Give Up Part D What is the height of the final image? Express your answer to two significant figures and include the appropriate units. f10 cm-30 cm 1.0 cm 1.539 cm 5.0 cm 5.0 cm lens mirror Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining 4:30 PM Type here to search 7/23/2017

Explanation / Answer

for first lens:

f1 = 10 cm

Object distance, p1 = 5 cm

Applying lens equation,

1/f1 = 1 / p1 + 1 / q1

1/10 = 1/5 + 1/q1

q1 = - 10 cm

so image is formed 10 cm left to lens

now this image will act as object for mirror.

p2 = 15 cm

f2 = - 30 cm

1/(-30) = 1/15 + 1/q2

q2 = - 10 cm behind the mirror.

now rays will pass the lens once again,

p3 = 15 cm

so, 1/10 = 1/15 + 1/q3

q3 = 30 cm left to first lens.

m = (-q1/p1) (-q2/p2)(-q3/p3)

= - (-10/5) (-10/15) (30 / 15)

= - 2.67

(A) image is virtual.

m is negative so image is inverted.

Ans: Virtual, Inverted

(B) to theleft of lens

(C) 30 cm

(d) m = hi / ho

2.67 = hi / 1

hi = 2.7 cm

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