JAMB - Physics (2023 - No. 35)
An object is placed 35 cm away from a convex mirror with a focal length of magnitude 15 cm. What is the location of the image?
26.25 cm behind the mirror
10.5 cm behind the mirror
26.25 cm in front of the mirror
10.5 cm in front of the mirror
Selitys
f = 15cm (diverging mirror); u = -35cm(because object is in front of the mirror) v=?
⇒\(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{u}\)
⇒ \(\frac{1}{15}\) = \(\frac{1}{v}\) - \(\frac{1}{35}\)
⇒ \(\frac{1}{15}\) + \(\frac{1}{35}\) = \(\frac{1}{v}\)
⇒ \(\frac{105}{10}\) = \(\frac{1}{v}\)
v = \(\frac{105}{10}\) = 10.5cm
∴The image is 10.5cm behind the mirror, since v is positive.
⇒\(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{u}\)
⇒ \(\frac{1}{15}\) = \(\frac{1}{v}\) - \(\frac{1}{35}\)
⇒ \(\frac{1}{15}\) + \(\frac{1}{35}\) = \(\frac{1}{v}\)
⇒ \(\frac{105}{10}\) = \(\frac{1}{v}\)
v = \(\frac{105}{10}\) = 10.5cm
∴The image is 10.5cm behind the mirror, since v is positive.