An object 5.0 cm is length is placed at 20 cm in front of convex mirror of radius curvature 30 cm .find the position of image it's nature and size
Answers
object hieght(ho) -5cm
object distance(u)-20cm
focal length (f)- radius of curvature /2
focal length - 30 /2
focal length -15
image distance(v)
1/f=1/v+1/u
1/15=1/v+1/20
1/v=1/15+1/20
1/v=20+15/15×20
1/v=35/300
v=8.5
Hey Mate ❤
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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