Question

7. An object is placed at a distance of 100 cm from a converging lens of focal length
40cm. What is the nature and position of the image?​

Answer 1

Answer:

-0.67

Explanation:

object distance, u=-100 cm

Focal length, f=40 cm

Using the lens formula,

1/v - 1/u=1/f

or 1/v =1/f +1/u

=1/40 -1/100

or v=+ 66.7 cm

m =v/u

= (200/3)/-100

=-0.67

Answer 2

Answer:

The position of image is 66.66cm on the other side of optical centre

and magnification is 0.666

Explanation:

Let us consider

• The object distance be u

• The image distance be v

• The focal length is f

To find

★The nature and position of the image

Formula to be used

★1/v - 1/u = 1/f

★magnification , m = h'/h = v/u

Given data

u = - 100cm

v = ?

f = 40cm

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \implies \frac{1}{ v} - \frac{1}{ - 100} = \frac{1}{40} \\ \implies \frac{1}{v} + \frac{1}{100} = \frac{1}{40} \\ \implies \frac{1}{v} = \frac{1}{40} - \frac{1}{100} \\ \implies \frac{1}{v} = \frac{5 - 3}{200} \\ \implies v = \frac{200}{3} \\ \implies v \: = 66.66$

So the image distance , v = + 66.66cm

•The positive sign shows that the image is formed at a distance 66.66cm on the side of the optical centre .

Now magnification

$m = \frac{v}{u} \\ \implies m = \frac{66.66}{ - 100} \\ \implies m = - 0.666$

The lens magnifies the image and is -0.666

•The negative sign of m shows image is inverted and real .