(a) Find the nature, position and magnification of the images formed by a convex lens of focal length 0.20 m if the object is placed at a distance of :
(i) 0.50 m
(ii) 0.25 m
(iii) 0.15 m
(b) Which of the above cases represents the use of convex lens in a film projector, in a camera, and as a magnifying glass?
(a) f = 0.20 m
(i) u = 0.50 m
\(\frac{1}{v}\) - \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{v}\) - \(\frac{1}{-0.50}\) = \(\frac{1}{0.20}\)
\(\frac{1}{v}\) = \(\frac{1}{0.20}\) - \(\frac{1}{0.50}\)
v = 0.33 m
Image is formed 0.33 m behind the lens.
m = \(\frac{v}{u}\) = \(\frac{0.33}{-0.50}\)
= -0.66
Image is real and inverted.
(ii) u = 0.25 m
\(\frac{1}{v}\) - \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{v}\) - \(\frac{1}{-0.25}\) = \(\frac{1}{0.20}\)
\(\frac{1}{v}\) = -\(\frac{1}{0.20}\) - \(\frac{1}{0.25}\)
v = 1 m
Image is formed 1 m in behind the lens.
m = \(\frac{v}{u}\) = \(\frac{1}{-0.25}\)
= -4
Image is real and inverted.
(iii) u = 0.15 m
\(\frac{1}{v}\) - \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{v}\) - \(\frac{1}{-0.15}\) = \(\frac{1}{0.20}\)
\(\frac{1}{v}\) = -\(\frac{1}{0.20}\) - \(\frac{1}{0.15}\)
v = -0.60 m
Image is formed 0.60 m in front of the lens.
m = \(\frac{v}{u}\) = \(\frac{-0.6}{-0.15}\) = 4
Image is virtual and erect.
(b) Film projector: Case (ii)
Camera: Case (i)
Magnifying glass: Case (iii)