Question

An object 2 cm height is placed 20 cm in front of convex lens of focal length 10cm.Find a distance from the lens at which a screen should be placed in order to obtained a sharp image. What will be size and nature of image formed​

Answer 1

Answer :-

The screen should be placed at a distance of 20 cm from the convex lens .

Nature of image : Real, inverted .

Size : Same as object .

Explanation :-

We have :-

→ Height of object (hₒ) = 2 cm

→ Distance of object (u) = 20 cm

→ Focal length of lens (f) = 10 cm

______________________________

According to sign convention, we have :-

• u = -20 cm

• f = + 10 cm

Let's calculate the required distance of the screen by using the lens formula .

1/v - 1/u = 1/f

⇒ 1/v = 1/f + 1/u

⇒ 1/v = 1/10 + 1/(-20)

⇒ 1/v = 1/10 - 1/20

⇒ 1/v = (2 - 1)/20

⇒ 1/v = 1/20

⇒ v = 20 cm

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Now, let's calculate the size of the image .

m = v/u = hᵢ/hₒ

⇒ v/u = hᵢ/hₒ

⇒ 20/(-20) = hᵢ/2

⇒ -1 = hᵢ/2

⇒ hᵢ = 2(-1)

⇒ hᵢ = -2 cm

[Here, -ve sign shows inverted image.]

Answer 2

⚘ Question :-

• An object 2 cm height is placed 20 cm in front of convex lens of focal length 10 cm. Find a distance from the lens at which a screen should be placed in order to obtained a sharp image. What will be size and nature of image formed?

⚘ Answer :-

• Distance from the lens at which a screen should be placed is 20 cm.
• Size of image is 2 cm.
• Nature of image is real, inverted.

Explanation:

⚘ Given :-

• Height of object (h) = 2 cm
• Distance of object (u) = -20 cm
• Focal length of lens (f) = +10 cm

⚘ To Find :-

• Distance from lens at which screen should be placed (v)?
• Size of image (h')?
• Nature of image?

⚘ Solution :-

• Firstly let's calculate distance from lens at which screen should be placed (v) by using len's formula ::

We know that,

$\leadsto{\large{\boxed{\sf{\red{\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}}}}}$

Where,

• v is distance of screen
• u is distance of object
• f is focal length of lens

We have,

• u = -20 cm
• f = 10 cm
• v = ?

According to the question by using the formula we get,

➨ $\sf \dfrac{1}{v} - \dfrac{1}{-20} = \dfrac{1}{10}$

➨ $\sf \dfrac{1}{v} = \dfrac{1}{10} + \dfrac{1}{-20}$

➨ $\sf \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{20}$

➨ $\sf \dfrac{1}{v} = \dfrac{2 - 1}{20}$

➨ $\sf \dfrac{1}{v} = \dfrac{1}{20}$

By doing cross multiplication we get,

➨ $\bf\purple{v = 20}$

∴ Distance from lens at which screen should be placed is 20 cm.

• Now, let's calculate size of the image (h') by using magnification formula ::

We know that,

$\leadsto{\large{\boxed{\sf{\blue{m = \dfrac{v}{u} = \dfrac{h'}{h}}}}}}$

We have,

• v = 20 cm
• u = -20 cm
• h = 2 cm
• h' = ?

According to the question by using the formula we get,

➨ $\sf \dfrac{v}{u} = \dfrac{h'}{h}$

➨ $\sf {\cancel{\dfrac{20}{-20}}} = \dfrac{h'}{2}$

➨ $\sf \dfrac{1}{-1} = \dfrac{h'}{2}$

➨ $\sf -1 = \dfrac{h'}{2}$

➨ $\sf h' = -1\:\times\:2$

➨ $\bf\pink{h' = -2}$

∴ Size of image is 2 cm.

Now,

• As sign is negative therefore image is real, inverted.

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