Science, asked by sharmakashish6995, 1 day ago

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 2 screen. After that he moves gradually the flame towards the lens and each time focuses its image on the screen.

Answers

Answered by mksharma1248
0

Answer:

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

Answered by krativarshney211
0

Explanation:

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

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