Question

an object 20 cm in length is placed at the distance of 400 cm in front of a convex mirror of radius of curvature 50
cm. find the position of image

2

Answer 1

Step-by-step explanation:

Height of object=5cm

f=15 cm

u=-20cm

therefore,

1/v=1/f-1/u

=1/15+1/20

=7/60

so v =60/7

therefore height of image=60/7/20=3/7

so the image is virtual,erect and smaller in size by 3/7th of its original size

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Answer 2

$\underline {\underline{ \huge{ \tt{ \purple{SOLution}}}}}$

$\pmb{Radius \: of \: curvature (R) = 30 \: cm}$

$\pmb{f = R/2 = 30/2 = 15 cm}$

➤ u = -20 cm,

➤ h= 5 cm.

$\boxed{ \pmb{ \red{1/v +1/u = 1/f}}}$

$\pmb➤ \: \pmb{1/v = 1/15+ 1/20 = 7/60}$

$➤ \: \pmb{v = 60/7 = 8.6 cm.}$

Image is virtual and erect and formed behind the mirror.

$➤ \: \pmb{hi/h0= v/u}$

$➤ \: \pmb{hi/5= 8.6/20}$

$➤ \: \pmb{hi = 2.2 cm.}$

$\boxed{ \pmb{ \blue{Size \: of \: image \: is \: 2.2 \: cm}}}$