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A square of side 3cm is located at a distance of 25cm from a concave mirror of the focal length of 10cm. The center of the square is at the axis of the mirror and the plane is normal to the axis of the mirror. The area enclosed by the image of the square is


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Solution

Step 1: Given data

Height of the object: h=3cm
Width of the object: bo=3cm
Distance of object from mirror: u=-25cm
Focal length: f=-10cm
Let distance of image from mirror be v

It is given that the rods of the mirror is normal to the square and thus the mirror is parallel to the square.

Step 2: Formula used

Mirror formula is given as,

1v+1u=1f

The magnification produced by a mirror,
m=hih=-vu
where hi is the height of the image and ho is the height of the object

Step 3: Solution

Substituting the values in the mirror formula,

1v+1u=1f

1v+1-25=1-101v=125-110v=-503cm

Therefore magnification is,


m=hih=-vu

hi3=--503-25hi=-2cm

The negative sign of the height indicates that the image is inverted

Similarly, bi=-2cm

Therefore, the area of the image formed

Ai=hibiAi=2×2Ai=4cm2

Hence, the area enclosed by the image of the square is 4cm2.


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