A CONCAVE MIRROR PRODUCE a real image of half the size of the object placed at 60 cm in front of it .where should the object be placed to obtain a virtual image of double size of the object?
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This problem makes use of the relationships found in the mirror equation and magnification equation.
1f=1d0+1di1f=1d0+1di
M=hih0=−did0M=hih0=−did0
Since the image is twice as large as the object we know that we can use the magnification equation to tell us where the image is.
hi=2h0hi=2h0
−did0=2−did0=2
di=−2d0di=−2d0
=−69cm=−60cm
The sign change indicates the image is formed 69cm60cm behind the mirror and is virtual.
1f=1d0+1di1f=1d0+1di
M=hih0=−did0M=hih0=−did0
Since the image is twice as large as the object we know that we can use the magnification equation to tell us where the image is.
hi=2h0hi=2h0
−did0=2−did0=2
di=−2d0di=−2d0
=−69cm=−60cm
The sign change indicates the image is formed 69cm60cm behind the mirror and is virtual.
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