Question

An object is placed at a distance of 25 cm away from concave lens of focal length 15cm. fine

a) Image distance

b) Nature of Image

c) Magnification.​

Answer 1

Given:

The distance of the object(u) = 25 cm.

Focal length (f) = 15 cm.

To find:

• The position of the image.

• The nature of the image.
• The magnificent of the image.

Solution:

Let, the distance of the image from the lens be, 'v'.

The magnificent of the image be, 'm'.

The size of the image be, 'I'.

f= 15 cm

u = 10 cm

From the lens formula we get that,

$\dfrac{1}{f} = (\dfrac{1}{u} - \dfrac{1}{v})$

$\\ or, \dfrac{1}{15} = (\dfrac{1}{25}-\dfrac{1}{v})$

$or, (\dfrac{1}{25} - \dfrac{1}{v}) = \dfrac{1}{15}$

$or, (-\dfrac{1}{v})= ( \dfrac{1}{15} - \dfrac{1}{25})$

$or, (-\dfrac{1}{v}) = (\dfrac{5-3}{75})$

$or, (-\dfrac{1}{v})= \dfrac{2}{75}$

$or, \dfrac{1}{v} = \dfrac{-2}{75}$

$or, v = - \dfrac{25}{2}$

or, v = -12.5 cm.

∴ The image has formed at a distance of 12.5 cm behind the lens.

From the formula we get that,

$m = \dfrac{-v}{u}$

$or, m = \dfrac{-(-12.5)}{25}$

$or, m = \dfrac{12.5}{25}$

or, m = 0.5 [Magnificent is a ratio so, it has no unit.]

∴The magnificent  of the image is  0.5.

Since, the magnificent of the image is positive so, the image is virtual, erect,  and upright.

Answer:

a)The image has formed 12.5 cm behind the mirror.              

b)The image is virtual, erect and upright.

c)The magnificent of the image is 0.5.

Answer 2

Step 1: Given data

object distance, $u=25cm$

focal length, $f=15cm$

image distance, $v=?$

nature of image $=?$

magnification, $m=?$

Step 2: Calculating the image distance

Using the lens formula,

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

$\frac{1}{15}=\frac{1}{25} -\frac{1}{v}$

$\frac{1}{v}=\frac{1}{25} -\frac{1}{15}$

$\frac{1}{v}=\frac{3-5}{75}=\frac{-2}{75}$

$v=\frac{-75}{2}=-37.5cm$

Step 3: Calculating the magnification

We know,

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

$m=\frac{-(-37.5)}{25}=1.5$

Hence, the image distance is $37.5cm$ behind the lens, the image is real, inverted and magnified, and its magnification is $1.5$.

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