Question

A student has focused on the screen the image of a 2cm high candle
flame placed at a distance of 24cm from a convex lens of focal length
16cm. Use lens formula to calculate the distance of the screen from the
lens and length of the image formed?

Answer 1

$\huge\tt{\bold{\purple{\underbrace{\red{\mathfrak{ɾҽզմíɾҽժ⠀αղsաҽɾ♡༄}}}}}}$

$\huge{\fbox{\pink{Given:-}}}$

• h = 2 m = 200 cm
• u = -24 cm
• f = 16 cm

⠀

$\huge{\fbox{\pink{Solution:-}}}$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \\ \implies \frac{1}{v} - \frac{1}{ (- 24)} = \frac{1}{16} \\ \\ \implies \frac{1}{v} + \frac{1}{24} = \frac{1}{16} \\ \\ \implies \frac{1}{v} = \frac{1}{16} - \frac{1}{24} \\ \\ \implies \frac{1}{v} = \frac{(3 - 2)}{48} \\ \\ \implies \frac{1}{v} = \frac{1}{48}$

$\huge{\green{Distance ⠀from⠀ ⠀ screen = 48 cm.}}$

$\frac{h \: dash}{h} = \frac{v}{u} \\ \\ \implies \frac{h \: dash}{200} = \frac{48}{ - 24} \\ \\ \implies h \: dash = - 2 \times 200 \\ \\ \implies h \: dash = - 400$

$\huge{\purple{Height⠀ of ⠀image:- 400 cm.}}$

$\huge{\orange{Image⠀ is ⠀inverted.}}$

Answer 2

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.

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.