Question

An object of height 4 cm is placed on the principal
axis at a distance of 20 cm in front of a convex

lens of focal length 30 cm. Find the distance of
image from the lens, its height and nature​

Answer 1

Answer:

From the lens formula,

=  1/v  −  1/u   = 1/f

⟹  1/f =  {1/24 }  − {1/−12}=1/8 ⟹f=8

When u=−12, the object is placed between F and 2F and the image is formed beyond 2F on the other side. This is a enlarged inverted image. If the object is moved away from the lens, the image moves towards the lens. So the screen should be moved towards the lens.

Magnification=v/u  =  24/-12 / =−2

When denominator increases, v decreases and hence magnification decreases.

Answer 2

Working out:

The above question is related to the image formation and calculation is based on lens formula used for Concave and convex lens. The lens formula is given by:

$\large{ \cdot{ \boxed{ \sf{ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} }}}}$

Here,

• v = image distance
• u = object distance and
• f = focal length of the lens.

In the question,

• Object distance 'u' = -20 cm
• Focal length 'f' = 30 cm

By using the lens formula,

⇛ 1/v - 1/u = 1/f

Plugging the given values,

⇛ 1/v - 1/-20 = 1/30

⇛ 1/v = 1/30 - 1/20

⇛ 1/v = 2-3/60

⇛ v = -60 cm

• Image distance = -60 cm (Other side of the lens)

Finding magnification,

⇛ m = v/u

⇛ m = -60/-20

⇛ m = 3

Height of object

⇛ hi/ho = 3

⇛ hi/4 = 3

⇛ hi = 12 cm

• Height of the image = 12 cm

Required answer:

• Distance of image from lens = 60 cm
• Height of the image = 12 cm
• Nature : Real, inverted & magnified by 3 times

And we are done !!