CBSE Class 10 Science 2017 · Paper 31(B)/C · Question 22

Light - Reflection and Refraction 5 marks Medium

(a)A student has focused on the screen the image of a 2 cm high candle flame placed at a distance of 24 cm from a convex lens of focal length 16 cm. Use lens formula to calculate the distance of the screen from the lens and length of the image formed.
(b)If the distance between the lens and the flame is increased to 32 cm, where would the image be formed and what would be its size ? Give reason to support your answer.

Answer

(a) Screen 48 cm from the lens, image 4 cm long; (b) image at 32 cm ( $2 F_{2}$ ), 2 cm long, as the flame is at $2 F_{1}$ .

(a) With

u = - 24

cm and

f = + 16

cm,

\frac{1}{v} = \frac{1}{16} - \frac{1}{24} = \frac{1}{48}

, so

v = + 48

cm: the screen is

48

cm from the lens, on the other side from the flame.

m = \frac{v}{u} = \frac{48}{- 24} = - 2

, so

h^{'} = - 2 \times 2 = - 4

cm: the image is

4

cm long, inverted, real and magnified. (b) With

u = - 32

cm,

\frac{1}{v} = \frac{1}{16} - \frac{1}{32} = \frac{1}{32}

, so

v = + 32

cm: the image is formed

32

cm from the lens on the other side, at

2 F_{2}

m = \frac{32}{- 32} = - 1

, so

h^{'} = - 2

cm: the image is

2

cm long, the same size as the flame, inverted and real. Reason:

32

cm is twice the focal length, so the flame is at

2 F_{1}

, and a convex lens forms the image of an object at

2 F_{1}

2 F_{2}

, real, inverted and of the same size.

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