An object of height 2.2 cm is placed 5.1 cm in front of a diverging lens of focal length 19 cm and is observed through the lens from the other side 50% part (a) how many centimeters beyond the lens does the observer locate the object's image? grade summa deductions potential ry 0% 100% coso asin0a sin submissions attempts remaining: j5 (5% per attempt) detailed view cotana acoso acotan0sinh0 atan cosh0 tanh cotanho degrees radians end deli clear submit hint i give up! hints: deduction per hint. hints remaining: feedback: 296 deduction per feedback. là 50% part (b) what is the height of the image, in centimeters?

a)   i = -4.02 cm , b)     h’= 1,576 cm

Explanation:

a) The constructor equation is

              1 / f = 1 / i + 1 / o

Where f is the focal length, i and o are the distance to the image and the object

Let's clear the distance to the image

             1 / i = 1 / f - 1 / o

             1 / i = 1 / -19 - 1 / 5.1

             1 / i = 0.2487

              i = -4.02 cm

b) let's use the expression of magnification

              m = h’/ h = - i / o

              h’= - h i / o

              h’= 2.2 4.02 /5.1

              h’= 1,576 cm

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