Question

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Answer 1

Answer:

1.32/2=16 cm

2.

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Answer 2

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

• 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