2. If you place an object very far from themirror, what is the location of the i

Question

2. If you place an object very far from themirror, what is the location of the image?Answer: : 37.3cm
3. What is the magnification of theimage?Answer: :0

5. What is the magnification of theimage?Answer: :-1.7

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NEED ONLY HELP WITH QUESTIONS 6and 7 thanks

6. If a diverging lens of focal length -67.0cm is then placed 27.0 cm from the first lens on the opposite sideof the lens from the object, what is the location of the newimage?

Answer:   ?????

Hint: If you know themagnification of the first lens and the magnification of the secondlens, you are just about there.
7. What is the magnification due to thetwo-lens system?

Answer: ????

NEED ONLY HELP WITH QUESTIONS 6and 7 thanks

Explanation / Answer

1.the focal length of the mirror is f = (r/2) r = 74.6 cm or f = (74.6/2) = 37.3 cm 2.the focal length of the mirror is (1/f) = (1/u) + (1/v) when u = then (1/f) = (1/) + (1/v) or (1/f) = (1/v) or v = f = 37.3 cm 3.the magnification of the image is m = (v/u) = (37.3/) = 0 4.we know that (1/f) = (1/u) + (1/v) v is the image distance or (1/v) = (1/f) - (1/u) = (1/30.0) - (1/47.65) = (9.53 -6/285.9) = (3.53/285.9) or v = (285.9/3.53) = 81.0 cm 5.the magnification of the image is m = -(v/u) = -(81.0/47.65) = -1.7 6.let the location of the new image be v cm from the diverginglens therefore we get (1/f) = (1/u) + (1/v) or (1/v) = (1/f) - (1/u) f = -67.0 cm and u = 27.0 cm or (1/v) = -(1/67.0) - (1/27.0) = -[(27.0 + 67.0/1809] or v = -(1809/94) = -19.2 cm 7.the magnification due to the two-lens system m = (v/u) = (19.2/27.0) = 0.71

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