The negative sign tells us that when s_i and s_0 are positive, the image will be

Question

The negative sign tells us that when s_i and s_0 are positive, the image will be inverted. You have a converging lens that has a focal length of f = 30cm. Calculate the image distance s_i the magnification m and indicate whether the image is upright or inverted for each of the following object distances. What is the percent difference in st for an object located at s_0 = 10/and s_0 = infinity? What about s_0 = 100/and s_0 = infinity? When is it justified to consider at object to be located at "optical infinity"?

Explanation / Answer

formula = 1/v - 1/u = 1/ f

1) u= 1f = 30cm

so

1/ v -1/-30 = 1/ 30

v= infinite

m = v/ u = infinite

case 2 u = 3*30 = - 90 cm

from here

v = 45 cm real inverted m = v/u = 45 /-90 = - 0.5

case 3 u = -300 cm

v = 33.33 cm real inverted m = v/u = 33.33 / -300 = -0.11

case 4

u = - 100*30 = -3000 cm

v = 30.303 cm real inverted m = v/u = 30.303/ 3000 = 0.01

case 5

u = infinite

v = 30 cm m = 0 real inverted

case 6

(s10 -sinfini / s10) = (33.33 - 30 ) / 33.33 = 0.0999

% = 0.099 *100 = 9.99

part 2

% = ( 30.303 - 30.0 )*100 / 30.303 = 0.99

in case of 100f we can consider it as optical infinite

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