please cover all parts -0.39m) is placed to the left of a diverging lens dfs-0061 m). The candle is do-a 12 m to the (13%) Problem 5. left of the lens. A candle (h ▲ 25% Part (a) Write an express...

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please cover all parts

Answer 1

Answer #1

Part A ) Using thin lens equation

1/f=1/d_o +1/d_i

1/d_i=1/f-1/d_o =d_o -f/d_of

d_i =(d_o*f)/(d_o -f) answer.

part B) Here f=-0.061 m , d_o = 0.12 m

d_i =-0.12*0.061/(0.12 +0.061)

d_i = -0.040442 m

part c) Since d_i is negative so image is virtual.

part d) SINCE Magnification m=h_i/h₀ =-d_i/d_o

m=h_i/h₀ =- (-0.040442)/0.12

=0.337

h_i=0.337*0.39

=0.13143 m

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Answer 2

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