Question

The image obtained with a convex lens is erect and its height is four times the height of the object. If the focal length of the lens is 20 cm, calculate the object distance and the image distance.

Answer 1

Correct Question

The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, calculate the object distance and the image distance.

Given that, image obtained with a convex lens is erect and its height is four times the height of the object.

Image distance from the lens = v

Object distance from the lens = u

Height of image = hi

Height of object = ho

As per given condition,

⇒ v = 4u

Also given that, the focal length of the lens is 20 cm (f = 20 cm).

We have to find the object distance and the image distance from the lens.

Now, using Lens formula:

⇒ 1/f = 1/v - 1/u

⇒ 1/20 = 1/4u - 1/u

⇒ 1/20 = - 3/4u

⇒ 1/5 = -3/u

⇒ u = -15 cm

Now,

⇒ v = 4(-15)

⇒ v = -60 cm

Also,

Magification (m) = v/u = hi/ho

⇒ -60/(-15) = hi/ho

⇒ 4 = hi/ho

⇒ hi = 4ho

Height of image is four times the height of object.

Therefore,

Object distance (u) from the lens is -15 cm and Image distance (v) from the lens is -60 cm.

Answer 2

Given:-

The image obtained with a convex lens is erect.

Its height is four times the height of the object.

Focal length is 20 cm.

To find:-

The object distance and the image distance.

Solution:-

f = 20 cm

Using lens formula,

$\displaystyle{\frac{1}{v}-\frac{1}{u} = \frac{1}{f} }\\\\\displaystyle{\implies \frac{1}{20}=\frac{1}{v} -\frac{1}{u} }\\\\\displaystyle{\implies \frac{1}{v}=\frac{1}{20}+\frac{1}{u} }\\\\\displaystyle{\implies \frac{1}{v}=\frac{u+20}{20u} }\\\\\displaystyle{\implies v=\frac{20u}{20+u} }$                          ...(i)

We know,

$\displaystyle{\frac{v}{u}= \frac{h_i}{h_o} }\\\\\displaystyle{\implies \frac{20u}{20+u}*\frac{1}{u}=\frac{4h_o}{h_o} }\\\\\text{(It is given that the height of the image is four times the height of the object.)}\\\\\displaystyle{\frac{20}{20+u} =4}\\\\\displaystyle{\implies 5=20+u}\\\\\displaystyle{\boxed{\implies u=-15\ cm}}$

Now, putting the value of 'u' in eq.(i) :-

$\displaystyle{v=\frac{20u}{20+u} }\\\\\displaystyle{\implies v=\frac{20*(-15)}{20+(-15)} }\\\\\displaystyle{\implies v=\frac{-300}{5} }\\\\\displaystyle{\boxed{\implies v=-60\ cm}}$

∴Thus, the object distance is -15 cm and the image distance is -60 cm.