Question

An object is placed 65.5 cm from a screen. (a) Where should a converging lens of focal length 9.5 cm be placed to form a clear image on the screen? (Give your answer to at least one decimal place.) shorter distance cm from the screen farther distance cm from the screen (b) Find the magnification of the lens. magnification if placed at the shorter distance magnification if placed at the farther distance

Answer 1

(a) Let do = x then di = 65.5 - X where do is the object distance and di is the image distance.

1/f = 1/do + 1/di

1/9.5 = 1/X + 1/(65.5 - X)

9.5 = X * (65.5 - X)/65.5

622.25 = 65.5 X - X²

X² - 65.5 X + 622.25 = 0

solving the quadratic equation,

X = 53.97 or 11.53

for shorter distance from the screen :-
di = 11.53 cm

for farther distance from the screen :-

di = 53.97 cm
so the lens can be placed between the object and the screen
either at 11.53 cm from screen or 53.97 cm from screen .

(b)

M = - di/do
for shorter di
M = - 11.53/ 53.97 = -0.214
for longer di
M = - 52.97 / 11.53 = - 4.594
the negative means that the image is real and inverted .