Question

please answer the questions of physics class 10th.​

Answer 1

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We have been given that :-

• Height of object - 4cm
• Object distance - 16cm-u
• Image distance - 60cm - v
• Height of image - $h_i$ = ?
• Focal length - f = ?

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In this question we have been given that an object of particular height is kept at a particular distance from a concave mirror and the image forms at a particular distance from the mirror, size of image formed and focal length of mirror is not known we have to calculate it.

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We know according to mirror formula :-

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

By this expression we can say that :-

$f = \frac{uv}{u + v}$

After putting the know values we get :--

$f = \frac{ - 16 \times 60}{16 - 60}$

$f = \frac{ - 960 }{ - 44}$

$f = \cancel \frac{ - 960}{ - 44}$

$f = 21.8cm$

Therefore the focal length of mirror is 21.8cm.

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Now, we know that magnification is the ratio of image distance to object distance which is also equal to ratio of height of image to height of object. Represented by :-

$m = \frac{ - v}{u} = \frac{h_i}{h_o}$

After putting the known values we get :-

$\frac{60}{16} = \frac{h_i}{4}$

$h_i = \frac{60 \times 4}{16}$

$h_i = 60 \times \cancel \frac{4}{16}$

$h_i = \frac{60}{4} = \cancel \frac{60}{4}$

$h_i = 15cm$

Therefore the height of image formed is 15cm.

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BY SCIVIBHANSHU

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