An object is placed $15 cm$ from a thin convex lens on its principle axis and a virtual image of certain size is formed. Now the object is moved $10 cm$ away from the lens and real image of the same size as that of the previous) virtual image is formed. The focal length (in cm) of the lens is .

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Solution

Let the focal length of the lens is $f$ and magnification is $m$ .

Case-I: when object is at $15 cm$ , image is virtual.
$u = - 15 cm$
Using, $m = \frac{v}{u}$ , $v = - 15 m$

Lens formula, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{f} = \frac{1}{- 15 m} + \frac{1}{15}$

$\frac{15}{f} = 1 - \frac{1}{m}$ .....(1)

Case-II: when object further moved $10 cm$ away from lens, image is real (magnification is negative).
$u = - (15 + 10) = - 25 cm$
Using, $m = \frac{v}{u}$ , $v = 25 m$

Again by lens formula, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{f} = \frac{1}{25 m} + \frac{1}{25}$

or, $\frac{25}{f} = 1 + \frac{1}{m}$ .....(2)

Now, equation (1) + (2):
$\frac{40}{f} = 2 \Rightarrow f = 20 cm$

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Complete the following sentences :

(a) An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and ..........

(b) An object is placed at a distance 2f from a convex of lens of focal length f. The size of image formed is ............ that of the object.

(c) An object is placed at a distance 5 cm from a convex of focal length 10 cm. The image formed is virtual, upright and ............

An object is placed at a distance 40 cm from a convex lens of focal length 20 cm. The image formed is:

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