A student focuses the image of a candle flame, placed at about 2 m from a convex lens of focal length 10 cm, on a screen. After that, he moves gradually the flame towards the lens and each time focuses its image on the screen.
(a) In which direction does he move the lens to focus the flame on the screen?
(b) What happens to the size of the image of the flame formed on the screen?
(c) What difference is seen in the intensity (brightness) of the image of the flame on the screen?
(d)What is seen on the screen when the flame is very close (at about 5 cm) to the lens?
please help me. THANKS! :)
Answers
Answer:
Explanation:
A)
v
1
=
f
1
+
u
1
, u is negative. let it be -U.
=
f
1
−
u
1
U is decreased. 1/U increases. So RHS decreases. So
v
1
decreases.
So v increases. Thus the lens is moved away from screen, towards the object.
B) As v increases and u decreases, magnification increases. So image is magnified.
C) As the image is enlarged, the light rays are spread out. So the brightness or intensity of the image is reduced.
D) When the flame is at Focal length distance, the image is at Infinity. If the flame is closer than focal length, then a virtual erect image is formed.
v
1
=
f
1
+
u
1
=
10
1
−
5
1
=−
10
1
v = 10 cm, a virtual image is formed.
magnification =
u
v
=
−5
−10
=2 So image is erect.