Question

A concave mirror has Focal length of 18cm.If a object having height of 7cm ,is placed 27cm away and Infront of the mirror,then what will be the size and nature of the image?​

Answer 1

Answer:

Question

✳️A concave mirror has Focal length of 18cm.If a object having height of 7cm ,is placed 27cm away and Infront of the mirror,then what will be the size and nature of the image?

Solution ⬇️

✍️ Remember: Focal length will always be negative for concave mirror

✍️ Given:u=-27cm , F=-18cm , h=7cm

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✍️ Using formula

$= = > \frac{1}{v} + \frac{1}{u} = \frac{1}{f} ..mirror \: formula$

$= = > \frac{1}{v} + \frac{1}{ - 27} = \frac{1}{ - 18} ...substitute$

$= = > \frac{1}{v} = \frac{1}{27} - \frac{1}{ - 18} ... \\ ( - 27 \: will \: turn \: 27)$

$= = > \frac{1}{v} = \frac{1}{ - 54}$

$= = > magnification \: = \frac{h2}{h1} = \frac{ - v}{u} = \frac{h2}{7.0} = \frac{ - (- 54)}{ - 27}$

$= = > \frac{h2}{7.0} = \frac{54}{ - 27} = 27 \times h2 = 54 \times 7.0 \\ h2= \frac{54 \times 7.0}{ - 27} = \frac{399}{ - 27} = - 14.77$

✍️ Therefore the image will be real and inverted and the size will be -14.77

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Extra Information

• Si unit of G = 6.673 x 10-¹¹ Nm²kg-²

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Extra formulae

$= > v = gt$

$= > s = \frac{1}{2} gt {}^{2}$

$= > v {}^{2} = 2gs$

Step-by-step explanation:

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Answer 2

$\sf \purple {Mirror \: Formula : } \begin{cases} \sf{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} } \end{cases}$

So, u = − 27cm

Concave mirror have negative focal length.

So, f = − 18 cm

Putting the values of u and f in the mirror formula

$\sf{ \frac{1}{v} + \frac{1}{ - 27} = \frac{1}{ - 18} }$

We get, v = 54cm

Also,

Magnification = $\sf{ \frac{h_2}{h_1} = \frac{ - v}{u} }$

$\longmapsto \sf{ \frac{h_2}{7.0} = \frac{ - 54}{ - 27} }$

$\sf{\pink{ \large{h_2 = - 14.0cm}}}$

$\large\textsf \green{Image will be real and inverted and} \\ \large\textsf \blue{ will be of size 14 cm.}$

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