Asked by Rajul
An object is placed at a distance of 1.5 m from a screen and a convex lens is interposed between them. The magnification produced is 4. What is the focal length of the lens.
Answers
Answered by
bobpursley
M= -di/do
solve for di.
then
1/f=1/di+1/do solve for f
solve for di.
then
1/f=1/di+1/do solve for f
Answered by
sahal
m = 4
= v/u =4
v = 4u
then,
1/f = 1/v - 1/u
substitute v with 4u,
1/f = 1/4u - 1/u
1/f = 1/4u - 4/4u
1/f = -3/4u
=f=-4u/3
total dist,
u + 4u = 1.5 * 100 (m-cm)
5u = 150
u = 150/5 = 30cm
substituting u in -4u/3
-4 * 30/3
-120/3 = -40cm
hope so you got it
this might be the ans
= v/u =4
v = 4u
then,
1/f = 1/v - 1/u
substitute v with 4u,
1/f = 1/4u - 1/u
1/f = 1/4u - 4/4u
1/f = -3/4u
=f=-4u/3
total dist,
u + 4u = 1.5 * 100 (m-cm)
5u = 150
u = 150/5 = 30cm
substituting u in -4u/3
-4 * 30/3
-120/3 = -40cm
hope so you got it
this might be the ans
Answered by
Anonymous
but the answer of this question in my guide is something else...not at all related to the one answered above