Question

 An object is placed at a distance of 40cm in front of a convex lens of focal length 20cm the image is formed at a distance of​

Answer 1

Given :

• Focal length (f) = 20 cm

• Object Distance (u) = 40 cm

To find :

The image distance (v).

Solution :

We know the lens formula b, i.e,

$\underline{\boxed{\bf{\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}}}}$

Where :-

• u = Object Distance
• v = Image Distance
• f = Focal Length

Now , using the lens formula and substituting the values in it, we get :-

$:\implies \bf{\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}}$

$:\implies \bf{\dfrac{1}{20} = \dfrac{1}{v} - \dfrac{1}{(-40)}}$

[Note :- In a convex mirror , the value of Object Distance (u) is always negative and the values of Focal length (f) and image distance (v) is always postive]

$:\implies \bf{\dfrac{1}{20} + \dfrac{1}{(- 40)} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{(- 2) + 1}{(- 40)} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{(- 1)}{(- 40)} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{(\not{-} 1)}{(\not{-} 40)} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{1}{40} = \dfrac{1}{v}}$

By Cross-Multiplication , we get :-

$:\implies \bf{v = 40}$

$\underline{\boxed{\therefore \bf{Image\:Distance\:(v) = 40\:cm}}}$

Hence, the image distance is 40 cm.

Answer 2

$\underline{\underline{\pink{\sf Given:}}}$

• Object Distance, u = 40cm
• Focal Length, f = 20cm

$\underline{\underline{\pink{\sf Find:}}}$

• Image Distance, v = ?

$\underline{\underline{\pink{\sf Solution:}}}$

we, know that

Lens Formula

$\underline{\boxed{ \sf\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} }}$

where,

• u = -40
• f = 20

So,

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

$\sf \to\dfrac{1}{v} - \dfrac{1}{( - 40)} = \dfrac{1}{20}$

$\sf \to\dfrac{1}{v} + \dfrac{1}{40} = \dfrac{1}{20}$

$\sf \to\dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{40}$

Take L.C.M

$\sf \to\dfrac{1}{v} = \dfrac{1 \times 2 - 1 \times 1}{40}$

$\sf \to\dfrac{1}{v} = \dfrac{ 2 - 1}{40}$

$\sf \to\dfrac{1}{v} = \dfrac{1}{40}$

Do, cross Multiplication

$\sf \to\dfrac{1}{v} = \dfrac{1}{40}$

$\sf \to 1 \times v = 40 \times 1$

$\sf \to v = 40cm$

Hence, distance of image will be 40cm