Question

an object is placed at a distance of 12 cm from a concave mirror of radius of curvature 16 cm , find the position of the image ?

no spams !

Answer 1

Answer :-

Image distance is -24 cm.

Explanation :-

Given :-

Object distance (u) = -12 cm

Radius of curvature (r) = -16 cm

To find :-

Position of the image/Image distance (v) = ?

Solution :-

Radius of curvature (r) = -16 cm

Focal length (f) = r/2

• -16/2

∴ Focal length is -8 cm.

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Using mirror formula :

1/v + 1/u = 1/f

1/v + 1/(-12) = 1/(-8)

1/v = 1/-8 - 1/-12

1/v = (1 × -3 - 1 × -2)/24

1/v = -1/24

v = -24

∴ Image distance (v) is -24 cm.

Answer 2

Given Information,

• Object Distance, u = - 12cm (according to sign convention)
• Radius of Curvature, R = - 16cm

Relationship b/w focal length and radius of curvature,

$\sf \: f = \dfrac{R}{2} \\ \\ \dashrightarrow \sf \: f = - \dfrac{16}{2} \\ \\ \dashrightarrow \boxed{ \boxed{\sf \: f = - 8 \: cm}}$

Using Mirror's formula,

$\sf \: \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} \\ \\ \dashrightarrow \sf \: \dfrac{1}{f} -\dfrac{1}{u} = \dfrac{1}{v} \\ \\ \dashrightarrow \sf \: \dfrac{1}{v} = \dfrac{1}{ - 8} - \dfrac{1}{ - 12} \\ \\ \dashrightarrow \sf \: \dfrac{1}{v} = \dfrac{ - 3 + 2}{24} \\ \\ \dashrightarrow \sf \: \dfrac{1}{v} = - \dfrac{1}{24} \\ \\ \dashrightarrow \boxed{ \boxed{ \sf \: v = - 24 \: cm}}$

Position of the image is - 24cm from the centre of curvature.