Question

Can someone help?

An object is placed on a metre scale at 8 cm mark was focused on a white screen placed at 92 cm mark, using a converging lens placed on the scale at 50 cm mark.(i)Find the focal length of the converging lens.(ii)Find the position of the image formed if the object is shifted towards the lens at a position of 29.0 cm.(iii)State the nature of the image formed if the object is further shifted towards the lens​

Answer 1

Answer:

i) 21 cm

ii) At infinity behind the lens.

iii) A virtual, upright, enlarged image behind the object

Explanation:

First identify,

object distance (u) = 42 cm (distance between  object and lens, 50 cm - 8 cm)

image distance (v) = 42 cm (distance between  image and lens, 92 cm - 50 cm)

The lens formula,

$\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

Then applying the new Cartesian sign convention to it,

$\frac{1}{v} +\frac{1}{u} =\frac{1}{f}$

Where f is (-), u is (+) and  v is (-) in  all 3  cases. (If not values with signs have to considered, this method that need will not arise)

Substituting values you get,

i) $\frac{1}{42} +\frac{1}{42} =\frac{1}{f}\\\frac{2}{42} =\frac{1}{f}$

f = 21 cm

ii) u =21 cm, f = 21 cm v = ?

Substituting in same equation$\frac{1}{v} =\frac{1}{21} =\frac{1}{21} \\\\\frac{1}{v} = 0$

  v ⇒ ∞ and image will form behind the lens

iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.