Question

THE FOCAL LENGTH OF A MIRROR IS 30CM FIND THE POSITION OF THE OBJECT IN FRONT OF THE MIRROR SO THE OBJECT THAT THE IMAGE IS 3TIMES THE SIZE OF THE OBJECT​

Answer 1

Answer:

f= -30 cm m= 3

U = ? V= v cm

\begin{gathered}m \: = \frac{ - v}{u} \\ 3 \: \: \: = \frac{ - v}{u} \\ 3u \: = - v\end{gathered}

m=

u

−v

3=

u

−v

3u=−v

\begin{gathered}\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{f} = \frac{v + u}{u \: v} \\ \frac{ - 1}{30} = \frac{ - 2}{ - 3u} \\ 3u = 30 \times - 2 \\ u = \frac{30 \times - 2}{3} \\ u \: = - 20cm\end{gathered}

f

1

=

u

1

+

v

1

f

1

=

uv

v+u

30

−1

=

−3u

−2

3u=30×−2

u=

3

30×−2

u=−20cm

Thus the object should placed 20 cm from the front of mirror.

Answer 2

Answer:

f is -ve. -4/3u = 1/(-30) => u = 40 cm that is 40 cm infront of the mirror. -2/3 u = -1/30 u = 20 cm that is 20 cm in front of mirror