Question

2. What power of a lens will change the reading
distance from 1 m to 0.25 m?

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Answer 1

Given:-

• Object distance ,u = 1m = -100 cm

• Image distance ,v = 0.25m = 25 cm

To Find:-

• Power of lens.

Solution:-

Firstly we calculate the focal length of the lens

Using Lens Formula

• 1/v - 1/u = 1/f

where,

v is the image distance

u is the object distance

f is the focal length

substitute the value we get

→ 1/25 - 1/(-100) = 1/f

→ 1/f = 1/25 +1/100

→ 1/f = 4+1/100

→ 1/f = 5/100

→ 1/f = 1/20

→ f = 20 cm

Now, calculating the Power of Lens.

• P = 1/f

where

P is the Power

f is the focal length in metre

20 cm = 20/100m = 1/5 m

Substitute the value we get

→ P = 1/1/5

→ P = 5/1

→ P = 5 D

Therefore, the power of the lens is 5 Dioptre.

Answer 2

Here, as per the provided question we are asked what is the power of lens…

So, let us start solving this question when the main informations are given here –

GIVEN :

• distance of the object = 1m
• distance of the image = 0.25m

TO FIND :

• power of a lens = ?

STEP-BY-STEP EXPLANATION :

Now, first of all we will have to change the distance of the object and distance of the image ( which is in metre into centimetre ) :

Changing the distance of the object and distance of the image (metre) → (centimetre) :

• distance of the object = 1m = -100cm

• distance of the image = 0.25m = 25cm

Therefore, here we will have to find the focal length (f) of the lens –

Formula Used –

• 1/v - 1/u = 1/f

[ substituting the values as per the formula ] :

$⟹ \: \frac{1}{25} - \frac{1}{( - 100)} = \frac{1}{ \rm {f}}$

$⟹ \: \frac{1}{ \rm {f}} = \frac{1}{25} + \frac{1}{100}$

$⟹ \frac{1}{ \rm {f}} = 4 \: + \frac{1}{100}$

$⟹ \: \frac{1}{ \rm {f}} = \frac{5}{100}$

$⟹ \: \frac{1}{ \rm {f}} = \frac{1}{20}$

$⟹ \rm {f} = 20 \rm {cm}$

Hence, we have got here focal length of the lens as 20cm.

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Now, on the basis of focal length of the lens we will find the power of lens here –

For finding the the focal length of the lens we will have to convert 20cm into metre = (20/100) → (1/5) metre.

$⟹ \rm {p} = \frac{1}{ \frac{1}{5} }$

$⟹ \rm {p} = \frac{5}{1}$

$⟹ \rm {p} = 5 \: \rm {D}$

Hence, power of the lens = 5D .