Question

an object 2.0 cm high Is kept at distance of 16 cm from a concave mirror which produces a real Image 3.0 cm high


1. what is the focal length of the mirror

2. find the position of the image ​

Answer 1

Answer:

-6.4cm

Explanation:

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Answer 2

height of the object, h1 = 2 cm

height of the image, h2 = -3 cm

object distance, u = -16cm

image distance = v

magnification =$\frac{h2}{h1} = \frac{-v}{u}$

⇒$\frac{-3}{2} = \frac{-v}{-16}$

⇒(-16)*(-3)=2*(-v)(−16)∗(−3)=2∗(−v)

⇒v = -$\frac{48}{2} =-24cmv$

=−24cm

From mirror formula

$\begin{gathered} \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \\ \\ \frac{1}{-16} + \frac{1}{-24} = \frac{1}{f} \\ \\ \frac{3+2}{-48} = \frac{1}{f} \\ \\ \frac{5}{-48} = \frac{1}{f} \\ \\ f= \frac{-48}{5} = -9.6cm \end{gathered}$

=−9.6cm

So focal length is 9.6cm and image distance is 24cm(to the left of mirror)