# Find the position of an object , which when placed in front of a concave mirror of focal length 10cm produces a virtual image , which is twice the size of the object.

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The Brainliest Answer!

h'=2h

where h is the object height and h' is the image height.

we know that,

m=h'/h

but h'=2h

or h'/h=2

so,m(magnification)=2

we know that m=h'/h=-v/u

where v is the image distance and u is the object distance.

so, m=-v/u

or, 2=-v/u

or, 2u=(-v)..............................(eq.1)

by mirror formula,

1/f=1/v+1/u

or, 1/v+1/u=1/f

1/v=1/f-1/u

1/v=u-f/fu

or, v=fu/u-f

=u(-10)/u-(-10)

therefore, v =(-10u)/u+10.......................(eq.2)

From eq.1,

2u=(-v)

2u=-[-10u/u+10]

=10u/u+10

2u=10u/u+10

or, 2u(u+10)=10u

or, u+10=10u/2u

or, u=5-10

or, u=(-5)

therefore, object distance is 5 cm to the left side of the mirror.

SAY THANKS PLZ....