Question

Find the power of convex lens which forms a real and inverted image of magnification – ½ of an object placed at a distance of 20 cm from its optical centre.

Answer 1

Answer :-

Power of the lens is + 15 D .

Explanation :-

We have :-

→ Distance of object (u) = 20 cm

→ Magnification (m) = -½

According to sign convention :-

• u = -20 cm

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Firstly, let's calculate the distance of the image from the lens .

m = v/u

⇒ -½ = v/(-20)

⇒ v = 20/2

⇒ v = 10 cm

Now, let's calculate focal length of the lens by using lens formula .

1/v - 1/u = 1/f

⇒ 1/10 - 1/(-20) = 1/f

⇒ 1/10 + 1/20 = 1/f

⇒ (2 + 1)/20 = 1/f

⇒ 3/20 = 1/f

⇒ 3f = 20

⇒ f = 20/3

⇒ f = 6.67 cm

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When, focal length of a lens is in 'cm' then relationship between power and focal length of the lens is given by :-

P = 100/f

⇒ P = 100/6.67

⇒ P = 14.99 ≈ 15 D

Answer 2

EXPLANATION:-

Using the formula first let's calculate the distance of the image from lens using:-

$\sf\bold{m=\frac{v}{u}}\\\\>-\frac{1}{2}=\frac{v}{-20}\\\\>v=10\;cm$

Now using the formula let's calculate the distance of image from lens:-

$\sf\bold{\frac{1}{v}-\frac{1}{u}=\frac{1}{f}}\\\\>\frac{1}{10}-\frac{1}{-20}=\frac{1}{f}\\\\>\frac{1}{10}+\frac{1}{20}=\frac{1}{f}\\\\>\frac{3}{20}=\frac{1}{f}\\\\>f=\frac{20}{3}$

Relationship between power and focal length of the lens ,when focal length is in 'cm' :-

$\sf\bold{p=\frac{100}{f}}\\\\p=\frac{100}{6.67}\\\\p=14.99\;D$

 or

p≈15 D