A bright object 50mm high stand on the axis of a concave mirror of focal length 100 mmand at a distance of 300mm from the concave mirror. How big will the image be?
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Answered by
82
f=100mm
u=300mm
h1=50mm
1/v+1/u=1/f
1/v+1/300=1/100
1/v=1/100 - 1/300
1/v= -1/200
v= -200
M=h2/h1= -v/u
h2=u/h1v
= 300/50(-200)
=-0.03mm
Hence, the height of image would be -0.03. The negative sign indicates that the image is on left side of concave mirror. And the image is inverted real and diminished.
HOPE THAT IT WILL HELP YOU!!!
u=300mm
h1=50mm
1/v+1/u=1/f
1/v+1/300=1/100
1/v=1/100 - 1/300
1/v= -1/200
v= -200
M=h2/h1= -v/u
h2=u/h1v
= 300/50(-200)
=-0.03mm
Hence, the height of image would be -0.03. The negative sign indicates that the image is on left side of concave mirror. And the image is inverted real and diminished.
HOPE THAT IT WILL HELP YOU!!!
Answered by
142
hope it will help you
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