an object 5.0 CM in length is placed at a distance of 20 CM in front of convex mirror of radius of curvature 30cm and the position and the image its nature and size
Answers
=30/2
=15cm
We know that, 1/f= 1/v+ 1/u
1/15 = 1/v -1/20
1/v =1/15 +1/20
1/v = 4 + 3/60
v = 60/7 cm
or
v = 8.57 cm
Size of the image is m = hi/ ho
also, m =- v/u
So, hi/ho =- v/ u
hi/5 = -60/-140
hi =3/7× 5
hi =15/7 cm
or
hi =2.1 cm
Nature of the image is virtual and erect
Size of the image is 2.1 cm
Position of the image is 8.57 cm
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Your Answer -
Given :
Object distance, u = -20 cm
Object height, h = 5 cm
Radius of curvature, R = 30 cm
Focal length, f = R/2 = 15 cm
☑ Mirror Formula :
According to the mirror formula,
1/v + 1/u = 1/f
1/v = 1/f - 1/u
1/v = 1/15 - ( - 1/20 )
1/v = 1/15 + 1/20
1/v = (3+4)/60
1/v = 7/60
v = 8.57 cm
➡ The positive value of v indicates that the image is formed behind the mirror.
☑ Magnification :
m = - v/u = - ( 8.57/-20) = 0.428
➡ The positive value of magnification indicates that the image formed is virtual.
Now again,
m = h'/h { h' is image height }
h' = m × h
h' = 0.428 × 5 = 2.14 cm
➡ The positive value of h' is indicates that the image formed is erect.
☑ Hence, the image formed is erect, virtual, and smaller in size.
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