Model Paper 1 — Question 4

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,

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

⇒ v = 3u .......... (1)

From lens formula,

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

On putting values

$\Rightarrow \dfrac{1}{3\text u}-\dfrac{1}{\text u}=\dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{3\text u}-\dfrac{3}{3\text u}=\dfrac{1}{15} \\[1em] \Rightarrow -\dfrac{2}{3\text u}=\dfrac{1}{15} \\[1em] \Rightarrow \text u = \dfrac{-15\times 2}{3}=-10 \text{ cm}$

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

(c) From (1)

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

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

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