Question

the focal length of convex lens is 18 cm and the size of the image is quarter of the object object is situated at the distance of

Answer 1

For lens, 1/v - 1/u = 1/f
For lens, v/u = Hi/Ho; where Hi is the image height, and Ho is the object height
Now, the ratio of image and object height is 1/4. Hence, v/u = 1/4 i.e. u = 4v.

Applying lens formula, 1/v-1/4v = 1/18 → 3/4v = 1/18 → v = 54/4
Now, u = 4v → u = 54/4×4 = 54 cm.

Answer 2

Let the height of the image and height of the Object be H₁ and H₀.
According to the Question,
 H₁ × 4 = H₀
∴ magnification  = 1/4 [∵ m = H₁/H₀]

Also, Magnification = v/u
where, v = Image distance, u = Object distance.
  1/4 = v/u
 u = 4v
This means that the object distance is four times the image distance.

Now,
Focal length = 18 cm
Image distance = v 
Object distance = u = 4v(negative)

Using the Mirror Formula,

   $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
 ⇒ $\frac{1}{18} = \frac{1}{v} + \frac{1}{-4u}$
⇒ v = 27/2 
  v = 13.5 cm.

Now, u = 4v = 4 × 13.5
= 54 cm.

Hence, Object is situated at the distance of 54 cm.


Hope it helps.