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Physics

An object is placed at a distance of 10 cm from a convex lens of focal length 20 cm.

(a) Find the position of the image.

(b) What is the nature of the image?

Refraction Lens

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Answer

Given,

  • Object distance (u) = - 10 cm
  • Focal Length (f) = + 20 cm

Let, image distance be 'v'.

From lens formula,

1f=1v−1u⇒1v=1f+1u⇒1v=120+1(−10)=120−110=1−2201v=−120⇒v=−201=−20 cm\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \\[1em] ⇒\dfrac{1}{\text v} = \dfrac{1}{\text f} + \dfrac{1}{\text u} \\[1em] ⇒ \dfrac{1}{\text v} = \dfrac{1}{20}+\dfrac{1}{(-10)}=\dfrac{1}{20}-\dfrac{1}{10}= \dfrac{1-2}{20}\\[1em] \dfrac{1}{\text v} = -\dfrac{1}{20} \\[1em] ⇒ \text v = -\dfrac{20}{1} = -20 \text { cm}

So, the image will form at a distance of 20 cm in front of the convex lens.

(b) As,

magnification=vu=−20−10=+ 2\text {magnification} = \dfrac {\text v}{\text u}= \dfrac {-20}{-10} = +\ 2

Here, sign of magnification is positive implying virtual and erect image formation and since magnification is 2 so size of the image is 2 times the size of the object.

Hence, formed images is virtual, erect and 2 times the size of the object.

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