Physics, asked by mrattitude50, 1 year ago

15 cm in front of a concave mirror as shown in the picture At a distance of 1 mm A square of the arm ABCD is kept. is . The focal length of the mirror is 10 cm is . The length of the perimeter (approximately) of the image of the square will be approx.

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Answered by sarthak720

1mm=.1cm,U=-15cm,f=-10cm

I/f =1/U+1/V= -30cm

HI/HO=-V/U=-2mm we need to neglect -ve sign now dist of image also matter -2mm belong y axis so we find x axis dist:- dV/dU=-(V/U)square =-4mm =-.4cm here we also neglect -ve sign now: Perimeter:-4*2+2*2=8+4=12mm or

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15 cm in front of a concave mirror as shown in the picture At a distance of 1 mm A square of the arm ABCD is kept. is . The focal length of the mirror is 10 cm is . The length of the perimeter (approximately) of the image of the square will be approx.​

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15 cm in front of a concave mirror as shown in the picture At a distance of 1 mm A square of the arm ABCD is kept. is . The focal length of the mirror is 10 cm is . The length of the perimeter (approximately) of the image of the square will be approx.