Question

please fast answers me ​

Answer 1

$\bigstar \large \underline \purple{ \pmb{ \sf 1st \: question }}$

Find the focal length of a convex mirror whose radius of curvature is 32 cm

$\bigstar \large \underline \purple{ \pmb{ \sf Concept... }}$

Radius of curvature of a convex mirror is +ve and focal length is half of it. So, we can use the following formula to find the value of focal length

$\underline{ \boxed {\sf focal \: length = \frac{radius \: of \: curvature}{2} }}$

$\bigstar \large \underline \purple{ \pmb{ \sf Solution... }}$

$: \implies \sf \: radius = \dfrac{ \cancel{32} ^{16} }{ \cancel2} = \red{16cm}$

hence, the required focal length is 16 cm.

$\bigstar \large \underline \purple{ \pmb{ \sf 2nd\: question }}$

A concave mirror produces 3 times enlarged image of an object kept 10 cm in front of it. Where is the image located ?

$\bigstar \large \underline \purple{ \pmb{ \sf Concept... }}$

in the question we are provided with the magnification of an image which is the ratio of image distance and object distance. Also, we are given the object distance

we are asked to find the image distance.

Object Distance is always -ve.

we can use the following formula to find the required value.

$\underline{ \boxed{ \sf magnification = \frac{ - v(image \: distance)}{u(object \: distance)} }}$

$\bigstar \large \underline \purple{ \pmb{ \sf Solution...}}$

$: \implies \sf 3= \dfrac{ \cancel - v}{ \cancel- 10}$

$: \implies \sf v = 3 \times 10 = \red{30cm}$

hence, the required image is formed at a distance of 30 cm behind the mirror.

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I hope it will help u to understand the concept well :D