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Question
an object of size 5cm is placed at 30cm infront of a concave mirror of focal length 20cm .At what distance should a screen be placed so that a sharp image formed on it? determine the nature and size.
an object of size 5cm is placed at 30cm infront of a concave mirror of focal length 20cm .At what distance should a screen be placed so that a sharp image formed on it? determine the nature and size.
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Solution
given that ;
Object size, h = + 5.0 cm;
Object distance, u = – 30.0 cm;
Focal length, f = –20 cm;
Image distance, v = ?
Image size, h′ = ?
but we have
Mirror formula is
(1/v) + (1/u) = 1/f
1/v = (1/f) − (1/u)
= (−1/20) + (1/30)
= −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/(– 30) = – 2
Height of the image h' = mh = (– 2)x5 = −10 cm
The image in inverted and enlarged.
given that ;
Object size, h = + 5.0 cm;
Object distance, u = – 30.0 cm;
Focal length, f = –20 cm;
Image distance, v = ?
Image size, h′ = ?
but we have
Mirror formula is
(1/v) + (1/u) = 1/f
1/v = (1/f) − (1/u)
= (−1/20) + (1/30)
= −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/(– 30) = – 2
Height of the image h' = mh = (– 2)x5 = −10 cm
The image in inverted and enlarged.
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