An object of size 7.0 cm is placed at a distance of 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and nature of the image.
Given: h1 = 7 cm, u = -27 cm, f = -18 cm, v = ?
Applying the formula \(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{u}\)
we get,
\(\frac{1}{-18}\) = \(\frac{1}{v}\) + \(\frac{1}{-27}\)
⇒ \(\frac{1}{-18}\) + \(\frac{1}{27}\) = \(\frac{1}{v}\)
⇒ \(\frac{-3 + 2}{54}\) = \(\frac{1}{v}\)
⇒ \(\frac{-1}{54}\) = \(\frac{1}{v}\)
⇒ v = -54 cm
Hence, the screen should be placed at a distance of 54 cm from the mirror.
Now, applying the formula m = -\(\frac{v}{u}\) = \(\frac{h_2}{h_1}\),
we get
⇒ \(\frac{-(-54)}{-27}\) = \(\frac{h_2}{7}\)
⇒ h2 = -2 x 7 = -14 cm
∴ Size of the image is 14 cm.
Nature of image: Negative sign shows that the image is inverted and enlarged.