Question

You place an object 23.6 cm in front of a diverging lens which has a focal length with a magnitude of 11.8 cm. Determine how far in front of the lens the object should be placed in order to produce an image that is reduced by a factor of 4.20

Answer 1

101.66 cm far in front of the lens the object is placed

Step-by-step explanation:

given data

distance = 23.6 cm

magnitude = 11.8 cm

factor = 4.20

to find out

how far in front of lens

solution

we will apply here lens formula that is

1/f = 1/p + 1/q ...................1

so here

q = f×p / ( p-f)

put here all value and get q

q = (-11.8 × 23.6) / 23.6 - (-11.8)

q = - −7.866 cm

q = −7.866 show image is on same side

and magnification here is

m = h1 / h2 = - q / p ......................2

here h1 = h2 (-q/p )

so h1 = - h2 ( −7.866 / 18 )

h1 = 0.437 h2

and we know new image factor is 4.20 time small

so

h3 / h2 = -q1 / p1

(h1/4.20) / h2 = -q1 / p1

(0.437 h2 / 4.20) / h2 = -q1 / p1

q1 = 0.1040 p1

so from equation 1

1/f = 1/p + 1/q

1/f = 1/p1 + 1/q1

- 1 /11.8 = 1/ p1 - 1 / 0.1040p1

so here

p1 = 101.66 cm

so 101.66 cm far in front of the lens the object is placed