an object 2cm high is placed at a distance of 16cm from a concave mirror which produces a real image 3cm high. then find length and position of the image
Answers
Answer:
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}
h1
h2
=
u
−v
⇒\frac{-3}{2} = \frac{-v}{-16}
2
−3
=
−16
−v
⇒(-16)*(-3)=2*(-v)(−16)∗(−3)=2∗(−v)
⇒v = - \frac{48}{2} =-24cmv=−
2
48
=−24cm
From mirror formula
\begin{lgathered}\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{lgathered}
u
1
+
v
1
=
f
1
−16
1
+
−24
1
=
f
1
−48
3+2
=
f
1
−48
5
=
f
1
f=
5
−48
=−9.6cm
So focal length is 9.6cm and image distance is 24cm(to the left of mirror)
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Answer:
length of a concave mirror is 8 and the position of the image is 24