Question

An object is placed at a distance of 10 cm in front of a concave lens of focal length 10 cm.
Find : (a) position of image
(b) The size of image in relation to the object.

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Answer 1

Information given to us:

• Distance at which an object is placed is 10 cm
• Focal length is 10 cm

Need to be calculated by us:

• Position and size of that image

Required formulas:

Lens formula:-

• 1/v - 1/u = 1/f

Linear magnification:-

• I / O = v / u

Performing calculations:

In the above given formula,

• distance of image is denoted by v
• distance of object is denoted by u
• focal length is denoted by f

The distance of object (u) if it is infront of lens it is always negative. The distance of image (v) would always be positive in the situation if it is real and had been formed behind the lens.

Here focal length (f) would also be negative.

Formula can be rewritten as,

➛ 1/v = 1/u + 1/f

Substituting the values,

➛ 1/v = 1/(-10) + 1/(-10)

➛ 1/v = - 2/10

➛ 1/v = 1/-5

We gets,

➛ v = -5 cm

Distance would be 5 cm as distance can't be negative.

We generally know that,

• The ratio of length of Image to the length of object it is known as linear magnification.

In the given above second formula,

• I denotes length of image
• O denotes length of object
• distance of object is denoted by u
• distance of image is denoted by v

Here we are finding out the size of image in relation to the object.

Substituting values,

➛ I / O = -5 / -10

Negative signs would be cancelled,

➛ I / O = 5 / 10

Cancelling them,

➛ I / O = 1 / 2

Answer:

• Position of image is 5 cm
• The size of image in relation to the object. is 1/2

Answer 2

Given:

An object is placed at a distance of 10 cm in front of a concave lens of focal length 10 cm.

To find:

• Position of image ?
• Size of image in relation to object ?

Calculation:

Applying LENS FORMULA:

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

$\implies \dfrac{1}{( - 10)} = \dfrac{1}{v} - \dfrac{1}{( - 10)}$

$\implies - \dfrac{1}{10} = \dfrac{1}{v} +\dfrac{1}{ 10}$

$\implies \dfrac{1}{v} = -\dfrac{1}{10} - \dfrac{1}{10}$

$\implies \dfrac{1}{v} = -\dfrac{2}{10}$

$\implies \: v = -5 cm$

So, image is formed at 5cm .

$\therefore \: magnification = \dfrac{v}{u}$

$\implies \: \dfrac{h_{i} }{h_{o}} = - 5/-10$

$\implies \: \dfrac{h_{i} }{h_{o}} = 0.5$

• So, image is half in size than object.