Question

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?

Answer 1

a) i = -4.02 cm , b) h’= 1,576 cm

Step-by-step 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