Question

Q. 16. The radius of curvature of a convex mirror is 30 cm. What
will be the size, position and nature of the image. If the object is placed
at 12 cm from the pole. Do the same calculation for a concave mirror?
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Answer 1

Explanation:

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Answer 2

Explanation:

$Focal \: length \: of \: convex \: mirror \: (ƒ) = \frac{R}{2}$

$∴ \: \: \: \: \: \: \: \: \: ƒ = \frac{30}{2} = + 15cm$

$Object \: distance \: \: (u) = - 12cm$

$∵ \: \: \: \: \: \: \: \: \: \frac{1}{ƒ} = \frac{1}{v} + \frac{1}{u}$

$⇒\: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{v} = \frac{1}{ƒ} - \frac{1}{u} \\ \\ \: \: \: \: \: \: = \frac{1}{15} - ( - \frac{1}{12} ) = \frac{1}{15} + \frac{1}{12}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{v} = \frac{4 + 5}{60} = \frac{9}{60}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: v = \frac{60}{9}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: v = \frac{20}{3} = 6.66cm$

$Linear \: magnification \: (m) = - \frac{v}{u}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: m = \frac{ - \frac{20}{3} }{ - 12}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: m = \frac{20}{3} \times \frac{1}{12} = \frac{5}{9}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: m = 0.5$

That is convex mirror will form smaller, virtual and erect image of the object at 6.66cm distance from it.