Question

An object of 5cm height is place at the distance of 10cm from a convex mirror of radios of 30cm find the nature and position and they size of the height of image
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Answer 1

$\bf \: \underline{Given} :$ ↴

$\sf \: \implies \: h_o = 5 \: cm$

$\sf \:\implies \: u = - 10 \: cm$

$\sf \:\implies \: R = 30 \: cm$

$\sf \: We \: know \: that,$

$\sf \:\implies \: f = \dfrac{R }{2}$

$\sf \: \implies \:f \: = \dfrac{30}{2}$

$\sf \: So, focal \: length \: (f) = 15 \: cm$

$\bf \: \underline{To \: find} :$ ↴

We need to find image distance (v), magnification (m) and height of image

$\: \red{ \implies \boxed{\sf \: \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }}$

$\sf \: \implies \: \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$

$\sf \: \implies \: \dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{ - 10}$

$\sf \: \implies \: \dfrac{1}{v} = \dfrac{1}{15} + \dfrac{1}{10}$

$\sf \: \implies \: \dfrac{1}{v} = \dfrac{2 + 3}{30}$

$\sf \: \implies \: \dfrac{1}{v} = \dfrac{5}{30}$

$\sf \: \implies \: \dfrac{1}{v} = \dfrac{\!\!\!\not5}{\!\!\!\not3\!\!\!\not0} = \dfrac{1}{6}$

$\sf \: \implies \: Image \: distance \: from \: mirror\: (v ) = + 6 \: cm$

The image is formed behind the image at the distance of 6 cm, hence image is vertual.

$\red{\implies \boxed{\sf \: m = \dfrac{ - v}{u} }}$

$\sf \: \implies m = \dfrac{ - 6}{-10}$

$\sf \: \implies \:M agnification \: ( m) = 0.6$

$\red{\implies \boxed{ \sf \: m = \dfrac{h_i }{h_o} }}$

$\sf \: \implies \dfrac{6}{10} \times 5 = h_i$

$\sf \implies \: \: \: Height \: of \: object \: (h_i )= + 3 \: cm$

The height of the image is 3 cm and the positive sign indicates that the image is erect.

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$\sf \: h_o = Height \: of \: object$

$\sf \: h_i = Height \: of \: image$

$\sf \: u = Object \: distance \: from \: mirror$

$\sf \: v = image \: distance \: from \: mirror$

$\sf \: R = Radius \: of \: curvature$

$\sf \: f = focal \: length \: of \: mirror$