Question

A lens forms the image of an object placed at a distance 15cm from it, at a distance 60 cm in front of it. Find :1)the focal length, 2)the magnification, and 3) the nature of image.

Answer 1

Answer:

Explanation: Here 'v' is taken negative because the image is forming in front of the lens and on the same side as of the object.

Answer 2

Answer:

The correct answer is

(i) The focal length of the lens is 20 cm.

(ii) The magnification exists 4.

(iii) The nature of the image exists erect, virtual, and magnified.

Explanation:

By using the lens formula

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

Given,

v = -60 cm

and u = -15 cm

Substituting the values in the given formula, we get,

$&\frac{1}{-60}-\frac{1}{-15}=\frac{1}{f}$

Simplifying the above equation as

$&-\frac{1}{60}+\frac{1}{15}=\frac{1}{f}$

$\\&\frac{1}{15}-\frac{1}{60}=\frac{1}{f}$

$\\&\frac{4-1}{60}=\frac{1}{f}$

$\\&\frac{3}{60}=\frac{1}{f}$

$\\&\Rightarrow f=\frac{60}{3}$

$\\&\Rightarrow f=20 \mathrm{~cm}\end{aligned}$

Therefore, the focal length of the lens exists 20 cm.

ii) As we know,

the formula for magnification of a lens exists

$m=\frac{v}{u}$

Given,

v = -60 cm

u = -15 cm

Substituting the values in the formula, we get,

$&m=\frac{-60}{-15}$

m = 4

Therefore, the magnification exists 4.

(iii) The nature of the image exists erect, virtual, and magnified.

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