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Physics

A lens of focal length 15 cm forms an erect image three times the size of the object on a screen.

(a) What kind of lens is this?

(b) Find the position of the object from the lens.

(c) Find the position of the screen from the lens.

Refraction Lens

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Answer

Given,

Focal length of the lens (f) = 15cm

Magnification of the lens (m) = +3 ∵ (image formed is virtual and erect)

(a) This is a convex lens because only convex lens forms a magnified image when the object is between the lens and the focus.

(b) Let,

Distance of object from the lens = u

Distance of image from the lens = v

As for lens,

m=vu⇒vu=3\text m = \dfrac{\text v}{\text u} \\[1em] \Rightarrow \dfrac{\text v}{\text u}=3

⇒ v = 3u ………. (1)

From lens formula,

1v−1u=1f\dfrac{1}{\text v}-\dfrac{1}{\text u}=\dfrac{1}{\text f}

On putting values

⇒13u−1u=115⇒13u−33u=115⇒−23u=115⇒u=−15×23=−10 cm\Rightarrow \dfrac{1}{3\text u}-\dfrac{1}{\text u}=\dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{3\text u}-\dfrac{3}{3\text u}=\dfrac{1}{15} \\[1em] \Rightarrow -\dfrac{2}{3\text u}=\dfrac{1}{15} \\[1em] \Rightarrow \text u = \dfrac{-15\times 2}{3}=-10 \text{ cm}

The object is placed 10 cm in front of the lens.

(c) From (1)

v=3u=3×(−10)=− 30 cm\text v = 3\text u =3\times (-10)=-\ 30 \text{ cm}

Distance of the screen from the lens = image distance (v) = -30 cm

The screen is placed 30 cm in front of the lens.

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