A 2cm high object is placed on the principle axis of concave mirror at a distance of 12cm from the pole. If the image is inverted, real and 5cm high, then the focal length of the mirror is.
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1/f = 1/u +1/v
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Answered by
43
Hey mate here is ur ans
we know that
image height /object height=image distance /object distance
=>5/2=x/12
=>2x=60
=>x=30cm
we know that in case of a concave mirror where a real image is formed
1/f=1/u+1/v
1/f=1/12+1/30=5+2/60=7/60
so f=60/7=8.57cm
the focal length of it is 8.57cm
we know that
image height /object height=image distance /object distance
=>5/2=x/12
=>2x=60
=>x=30cm
we know that in case of a concave mirror where a real image is formed
1/f=1/u+1/v
1/f=1/12+1/30=5+2/60=7/60
so f=60/7=8.57cm
the focal length of it is 8.57cm
Answered by
48
Height of object = +2cm
Height of image = -5cm
Distance of object (u) = -12cm
Magnifaction =Height of image /Height of object
M= (-5)/2= -2.5
And again
Magnifaction = -Image distance/object distance
-2.5= -v/-12
30= -v
v= -30cm
Using the mirror formula to find focal length
1/f=1/v+1/u
1/f=1/-30+1/-12
By lcm method
1/f= -2+-5/60
1/f= -7/60
cross multiple
-7f=60
f= -60/7 or -8.5 approx
I HOPE THIS WILL HELP YOU MARK ME BRAINLIEST
Height of image = -5cm
Distance of object (u) = -12cm
Magnifaction =Height of image /Height of object
M= (-5)/2= -2.5
And again
Magnifaction = -Image distance/object distance
-2.5= -v/-12
30= -v
v= -30cm
Using the mirror formula to find focal length
1/f=1/v+1/u
1/f=1/-30+1/-12
By lcm method
1/f= -2+-5/60
1/f= -7/60
cross multiple
-7f=60
f= -60/7 or -8.5 approx
I HOPE THIS WILL HELP YOU MARK ME BRAINLIEST
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