(a) An object 3 cm high is placed 24 cm away from a convex lens of focal length 8 cm. Find by calculations, the position, height and nature of the image.
(b) If the object is moved to a point only 3 cm away from the lens, what is the new position, height and nature of the image?
(c) Which of the above two cases illustrates the working of a magnifying glass?
(a) \(h_1\) = 3 cm
u = -24 cm
f = 8 cm
\(\frac{1}{v}\) - \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{v}\) - \(\frac{1}{-24}\) = \(\frac{1}{8}\)
\(\frac{1}{v}\) = \(\frac{1}{12}\)
v = 12 cm
Image is formed 12 cm behind the lens.
m = \(\frac{v}{u}\) = \(\frac{h_2}{h_1}\)
\(\frac{12}{-24}\) = \(\frac{h_2}{3}\)
\(h_2\) = -1.5 cm
Image is 1.5 cm high, real and inverted.
(b) u = -3 cm
\(h_1\) = 3 cm
f = 8 cm
\(\frac{1}{v}\) - \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{v}\) - \(\frac{1}{-3}\) = \(\frac{1}{8}\)
\(\frac{1}{v}\) = -\(\frac{5}{24}\)
v = -4.8 cm
Image is formed 4.8 cm in front of the lens.
m = \(\frac{v}{u}\) = \(\frac{h_2}{h_1}\)
\(\frac{-4.8}{-3}\) = \(\frac{h_2}{3}\)
\(h_2\) = 4.8 cm
Image is 4.8 cm high, virtual and erect.
(c) Case (b)