A lens of focal length 15.0 cm forms an upright image four times the size of an object. Calculate the distance of the image from the lens

Faroukalao

4 years ago

D... 45cm
Virtual, Upright and magnified shows a convex(converging) lens where the object if between F and the Pole
In this case V is negative

f = uv/u+v

M = v/u
4 = v/u
v = 4u

15 = -4u × u/(-4u + u)
15 = -4u²/-3u
15 = 4u/3
45 = 4u
u = 11.25cm

v = 4u
v = 45

... v = 45cm

Nnaaaaa

7 years ago

Upright image ia virtual and its negative

If its real
M=V/F -1
If it is virtual
M=V/F + 1

SO HERE ITS VIRTUAL WE USE THE SECOND FORMULA

M=4 f=15
4-1=v/15
V=15×3=45cm√√√

raph.Montella

9 years ago

Here is an explanation:

M=4, f=15cm, v= ?



M=(v/f) -1



4=(v/15) -1

4+1= v/15

5=v/15

15*5 =v

75=v

75.0cm = v

REF: Check: Essential Physics, page 227

Martins patrick

11 years ago

i disagree with dis ans i think its b.

fredo219

10 years ago

Wow!. Thanks my school

Holyfield22

10 years ago

Its 75.2cm