ab object 5 cm high is placed at a distance of 10 cm in front of a concave mirror of focal lengths 20 cm find position, nature and size of image
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HI,
Object-size, h = + 5.0 cm;
Object-distance, u = – 20.0 cm;
Focal length, f = –15.0 cm;
Image-distance, v = ?
Image-size, h′ = ?
Mirror formula is (1/v) + (1/u) = 1/f
1/v = (1/f) − (1/u)
= (−1/15) + (1/20)
= −1/60
v = − 60 cm.
The screen should be placed at 60 cm from the mirror. The image is real.
Magnification m = h'/h = − v/u = 60/(– 20) = – 3
Height of the image h' = mh = (– 3)x5 = − 15 cm
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Object-size, h = + 5.0 cm;
Object-distance, u = – 20.0 cm;
Focal length, f = –15.0 cm;
Image-distance, v = ?
Image-size, h′ = ?
Mirror formula is (1/v) + (1/u) = 1/f
1/v = (1/f) − (1/u)
= (−1/15) + (1/20)
= −1/60
v = − 60 cm.
The screen should be placed at 60 cm from the mirror. The image is real.
Magnification m = h'/h = − v/u = 60/(– 20) = – 3
Height of the image h' = mh = (– 3)x5 = − 15 cm
HOPE IT HELPS YOU
MARK IT AS AN BRAINLIESt THANKS
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