Question

answer all parts please

(8%) Problem 8: of the lens A candle (10-0 27 m) is placed to the left of a diverging lens-A.054 m). The candle is do-0.38 m to the left 25% Part (a) Write an expression for the image distance, a, Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: 5 do % per attempt) detailed view END No DEL CLEAR Submit Hint I give up! Hints: 1% deduction per hint. Hints remaining: 2 Feedback: 1% deduction per feedback 25% Part (b) Numerically, what is the image distance in meters? Δ 25% Part (c) Is this real or virtual? 25% Part (d) Numerically, what is the image height, hi? All content 2018 Expert TA LLC

Answer 1

image distance = v ,

object distance = u ,

focal length= f ,

magnification = m = v/u

Lens formula=>

A ) 1/di+1/do= 1/f

1/di = 1/f+1/do

di = ( do+f) / ( do-f)

B)

1/di- 1/-0.38 = 1/-0.054

di = -0.0472811 m

C) virtual

D) magnification = di/do = -0.0472811​/-0.38​ =​0.12442

so height = 0.12442*ho = 0.12442​*0.27 =0.0335934​ m