Physics, asked by nchinni908, 9 months ago

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

Answered by Anonymous
3

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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Answered by saibalghosh796
0

Answer:

length of a concave mirror is 8 and the position of the image is 24

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