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A small candle 2.5cm in size is placed at 27cm in front of a concave mirror of radius of curvature 36cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved ?

Posted by Sowmya V 3 years, 3 months ago

CBSE > Class 12 > Physics

2 answers

Anjan Karthi 3 years, 3 months ago

m = h'/h = v/u = 54/-27 = -2. Therefore, h' = -2 × 2.5 = -5 cm. Thus, the image is real, inverted and magnified. If u is reduced, when v should be increased.

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Anjan Karthi 3 years, 3 months ago

h = 2.5 cm, u = -27 cm, f = R/2 = -36/2 = -18 cm. Therefore, 1/f = 1/v + 1/u. 1/v = 1/f - 1/u. v = fu/f-u = 18 × 27/-18-(-27) = 18×27/9 = 54 cm.

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