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An object $1 m$ tall is placed on the principal axis of a convex lens and its $40 cm$ tall image is formed on the screen placed at a distance of $70 cm$ from the object. Focal length of the lens [[1]](with sign) $cm$ .

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Detailed Solution

An object

1 m

tall is placed on the principal axis of a Convex lens and its

40 cm

tall image is formed on the screen placed at a distance of

70 cm

from the object. Focal length of the lens

+ 14.3 cm

.
Since, the image is formed on the screen, so the image is real and inverted. Hence, the image height will be negative.
Given that,
Object height,

h

100 cm

Image height,

h'

- 40 cm

Let the object be kept at a distance

x

from the lens,

\Rightarrow

v

= +

(70 - x)

Focal length,

f

=?
Magnification,

m = \frac{h'}{h}

= \frac{v}{u}

\Rightarrow \frac{- 40}{100} = \frac{(70 - x)}{- x}

\Rightarrow

x = 50 cm

\Rightarrow u =

50 cm

Also,

v = 70 - x

\Rightarrow 70 - 50 = 20 cm .

\Rightarrow v = 20 cm

Substituting the values of u and v in the lens formula,

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

where

f

is the focal length,

u

is the object distance and

v

is the image distance, we get

\frac{1}{f} = \frac{1}{20} - \frac{1}{(- 50)}

f = \frac{100}{7} = + 14.3 cm

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