Question

Can someone help?

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.