A square wire of side `3.0 cm` is placed `25 cm` away from a concave mirror of focal length `10 cm`. What is the area enclosed by the image of the wire ? The centre of the wire is on the axis of the mirror, with its two sides normal to the axis.

Explanation

Here, `h_1 = 3 cm, u = -25 cm`,
`f = -10 cm, v = ?, m = ?`
As `(1)/(u)+(1)/(v)=(1)/(f)`
`:. (1)/(v)=(1)/(f)-(1)/(u)=(1)/(-10)+(1)/(25)=(-5 + 2)/(50)= - (3)/(50)`
`v = -50//3 cm`
As image is real,
`:. m = (-h_2)/(h_1) = (-v)/(-u) = (v)/(u)`
`:. (-h_2)/(3) = ((-50//3))/(-25) = (2)/(3)` or `h_2 = -2 cm`
(minus sign for inverted image)
Area of square wire `= (-2)^2 = 4 sq. cm`.