An object 0.05
M high is placed at a distance of 0.5 m
From concave mirror radius of curvature 0.2 M find the position nature and size of the image formed
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H=height of the object =5cm
Distance from the mirror =u=-50cm
R=0.2m=-20cm
So,focal length =20/2=10cm
By mirror formula
1/f=1/v+1/u
Put the values.
-1/20+1/50=1/v[Distance of image]
V=-3.3
So the image formed is real, inverted and small in size than the object.
Now H=5cm
Let him be the height of image
h/H=-v/u
h/5=3.3/20
h=3.3/4
Image size is small than the object
Distance from the mirror =u=-50cm
R=0.2m=-20cm
So,focal length =20/2=10cm
By mirror formula
1/f=1/v+1/u
Put the values.
-1/20+1/50=1/v[Distance of image]
V=-3.3
So the image formed is real, inverted and small in size than the object.
Now H=5cm
Let him be the height of image
h/H=-v/u
h/5=3.3/20
h=3.3/4
Image size is small than the object
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