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Q.

A converging lens is to project $4$ times enlarged image of a lamp on a wall at a distance of $10 m$ from the lamp. The focal length of the lens is.

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a

+ 2 m

b

+ 1.7 m

c

+ 70 cm

d

+ 1.6 m

answer is D.

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Detailed Solution

The Focal length of the lens,

f = + 1.6 m

Given,
Distance of the wall from the lamp,

| u + v | = 10 m

Size of the image is 4 times the size of the lamp, i.e.,
⇒

\frac{h ’}{h} = \frac{v}{u}

= 4

⇒

|v| = |4 u|

⇒

|u + v| = 10

\Rightarrow |u + 4 u| = 10

\Rightarrow 5 |u| = 10

⇒

u = 2 m

⇒

v = 4 |u|

= 4 \times 2

= 8 m

As per the sign convention followed,

u = - 2 m

and

v = + 8 m

Thus,
Formula,

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

= \frac{1}{+ 8} - \frac{1}{- 2}

⇒

\frac{1}{f} = \frac{1 + 4}{8}

⇒

\frac{1}{f} = \frac{5}{8}

⇒

f = \frac{8}{5}

= 1.6 m

.

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A converging lens is to project 4 times enlarged image of a lamp on a wall at a distance of 10 m from the lamp. The focal length of the lens is.