Question

A 3.00-cm tall object is placed 18.0 cm from the center of a converging lens with a 24.0 cm center of curvature.

a. Determine a scale, then use graph paper and color pencils or pens to neatly draw a complete ray diagram to scale.

b. Based on the scale of your ray diagram determine the image distance, height, and magnification.

c. Based on your drawing state whether the image is real or virtual, upright, or inverted.

d. Use the mirror and lens formulas to calculate the image distance, height, and magnification. Compare the calculated values to your ray diagram values and explain the differences.

Answer 1

Here is the ray diagram

the scale for horizontal axis and vertical axis

is 1 box = 3 cm

From the image, we can see that

image is real, inverted

image is on the other side of lens

image is at 12 boxes away means 12 * 3 = 36 cm

__________________

Now, Let's do calculations

we know

1/f = 1/p + 1q

where p is object distance and q is image distance

f = R / 2 = 24 / 2 = 12 cm

so,

1/12 = 1/18 + 1/q

1/q = 1/12 - 1/18

q = 36 cm

and

m = -q / p

m = -36 / 18

m = - 2

negative sign means image is real and inverted

2 means image is double the size of object which we can clearly see in ray diagram !!!

height of image = 6 cm