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Question

A square ABCD of side 1 mm is kept at distance 15 cm in front of the concave mirror as shown in figure. The focal length of the mirror is 10 cm. The length of the perimeter of its image will be

A
8 mm
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B
2 mm
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C
12 mm
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D
6 mm
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Solution

The correct option is C 12 mm
Here AB and CD will undergo longitudinal magnification and AD and BC will undergo lateral magnification.

For CD,

Magnification m=ffu=101510=2
Therefore CD=2×1=2 mm

Similarly, AB=2×1=2 mm

For BC,

vi=(vu)2v0

ΔvΔt=(vu)2ΔuΔt

ΔvΔu=(vu)2=f2(fu)2

[vu=m=ffu]

ΔvΔu=22=4

Therefore for BC,AD,

BC=AD=4×1=4 mm

Perimeter ABCD=2+2+4+4=12 mm

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