An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Given: h1 = 5.0 cm, u = -20 cm, R = 30 cm
∴ f = \(\frac{R}{2}\) = \(\frac{30}{2}\) = 15 cm, v = ?
Applying the formula \(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{u}\), we get
\(\frac{1}{15}\) = \(\frac{1}{v}\) + \(\frac{1}{-20}\)
⇒ \(\frac{1}{v}\) = \(\frac{1}{15}\) + \(\frac{1}{20}\)
= \(\frac{4+3}{60}\) = \(\frac{7}{60}\)
⇒ v = \(\frac{60}{7}\) = 8.57 cm
Hence, the image is formed at a distance of 8.57 cm behind the mirror. It must be virtual and erect.
Now, applying the formula \(\frac{-v}{u}\) = \(\frac{h_2}{h_1}\), we get
\(\frac{\frac{-60}{7}}{-20}\) = \(\frac{h_2}{5}\)
⇒ h2 = \(\frac{60 \times 5}{7 \times 20}\) = \(\frac{15}{7}\)
= 2.1 cm
This is the size of the erect image.