A concave mirror of focal length 10cm. A square wire of side 3cm is placed 23cm away from the mirror. The centre of the square lies on the principal axis. What is the area enclosed by the image?

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A concave mirror of focal length 10cm. A square wire of side 3cm is placed 23cm away from the mirror. The centre of the square lies on the principal axis. What is the area enclosed by the image?

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Answer 1 · 2017-09-19T11:08:10

We know that conjugate foci relation of spherical mirror Is as follows

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $1/v+1/u=1/f$ ,

Where

$v \to image distance = ?$ $v->"image distance"=?$

$u \to object distance = - 23 c m$ $u->"object distance"=-23cm$

$f \to focal length = - 10 c m$ $f->"focal length"=-10cm$

$\frac{1}{v} - \frac{1}{23} = - \frac{1}{10}$ $1/v-1/23=-1/10$

$\Rightarrow v = - \frac{230}{13}$ $=>v=-230/13$ cm

Magnification $m = \frac{v}{u} = \frac{\frac{230}{13}}{23} = \frac{10}{13}$ $m=v/u=(230/13)/23=10/13$

Length of each side of the square frame is $3$ $3$ cm

If the plane of the square wire frame is kept vertical with the principal axis the length of each side of its image will be $3 \times \frac{10}{13} = \frac{30}{13}$ $3xx10/13=30/13$ cm

Hence area enclosed by the image will be ${(\frac{30}{13})}^{2} c m^{2} \approx 5.33 c m^{2}$ $(30/13)^2cm^2~~5.33 cm^2$

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A concave mirror of focal length 10cm. A square wire of side 3cm is placed 23cm away from the mirror. The centre of the square lies on the principal axis. What is the area enclosed by the image?

1 Answer

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