Physics, asked by tuduraghunath8, 9 months ago

A child looks at a reflecting christmas tree ball that had a diameter of 9.0 cm and sees and image of her face that is half the real size . How far is the child 's face from the ball ?​

Answers

Answered by Uriyella
11

Question :–

A child looks at a reflecting christmas tree ball that had a diameter of 9.0 cm and sees and image of her face that is half the real size. How far is the child's face from the ball ?

Answer :–

The child face far from the ball is  \tt 2.25 cm

Given :–

  • Diameter = 9.0cm

To Find :–

  • How far is the child's face from the ball ?

Solution :–

Let,

p = distance of face to ball

q = distance of virtual image to ball

r = Radius of ball = 4.5cm

 \implies \tt r = \dfrac{9}{2} = 4.5cm

So,

 \implies \tt F = \dfrac{r}{2}

 \implies \tt \dfrac{4.5}{2}

 \implies \tt \dfrac{45}{2 \times 10}

 \implies \tt \dfrac{\cancel{45}}{\cancel{20}}

 \implies \tt \dfrac{9}{4}

 \implies \tt 2.25cm

 \tt M = 0.5

  • By Eq. 23.4, M  \sf = \dfrac{-d_{i}}{d_{o}}

\tt \implies 0.5 = \dfrac{-d_{I}}{d_{o}}

 \tt \implies d_{i} = -0.5d_{o}

  • By Eq. 23.3

 \implies \tt \frac{1}{d_{o} }  +  \frac{1}{ d_{i} }  =  \frac{1}{f}

 \implies \tt\frac{1}{ d_{o} }  +  \frac{1}{ { - 0.5}d_{o}}  =  \frac{1}{2.25}

 \implies \tt \frac{1}{d_{o}}   + (1 -  \frac{1}{0.5} ) =  \frac{1}{2.25}

 \implies \tt\frac{1}{ d_{o}}  \times  - 1 =  \frac{1}{2.25}

 \implies \tt d_{o} =  - 2.25cm

Hence,

The child face far from the ball is  \tt 2.25 cm

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