Question

An object is placed 30 cm away from a symmetric convex lens and an image two thirds of the size of the object is produced. The object is moved by a distance of 20 cm so as to get a magnified image. Now we get 

(a)  a real image of magnification 

(b)  a virtual image of magnification 5 

(c)  a real image at a distance of 40 cm 

(d)  a virtual image at a distance of 60 cm

Answer 1

Correct option  (d) a virtual image at a distance of 60 cm

Explanation:

Size of the image : size of the object ⇒  v = 2/3 u

Lens formula (real image) : 1/v - 1/u = 1/f 

⇒ 3/2u + 1/u = 1/f  

⇒ f = 2u/5 = 2 x 30 /5 = 12 cm

After moving the object by 20 cm, it is with in the focus of the lens. So it forms a virtual image on the same side as that of the object.

Lens formula (virtual image : 1/v - 1/u = 1/f 

⇒ - 1/v + 1/10 = 1/12 

⇒ 1/v = 1/60 

⇒ v = 60 cm