(13%) Problem 6: A candle (ho-0.26 m) is placed to the left of a diverging lens

Question

(13%) Problem 6: A candle (ho-0.26 m) is placed to the left of a diverging lens (f=-0.074 m). The candle is do-0.18 m to the left of the lens Randomized Variables ho= 0.26 m f=-0.074 m do= 0.18 m 25% Part (a) Write an expression for the image distance, di Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: 6 (390 per attempt) detailed view 0 0 BACKSPACE DHLCLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback 25% Part (b) Numerically, what is the image distance in meters? 25% Part (c) Is this real or virtual? 25% Part (d) Numerically, what is the image height, h

Explanation / Answer

a)

using the lens equation

1/f = 1/di + 1/do

di = f do /(do - f)

b)

di = (- 0.074) (0.18) /(0.18 - (- 0.074))

di = - 0.052 m

c)

since image distance is negative, the image is virtual

d)

hi = (- di/do) ho

hi = - (- 0.052/0.18) (0.26)

hi = 0.075

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