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An object is placed at (i) 10cm (ii) 5 cm in front of a concave mirror of radius of curvature 15cm. Find the position, nature and magnification of the image in each case. |
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Answer» SOLUTION :The focal length f = `(-15)/(2) ` cm = - 7.5 cm i. The object distance U = - 10cm . Then Eq. 6 GIVES `(1)/(v) + (1)/(-10) = (1)/(-7.5)or V = (10 xx 7.5)/(-2.5 ) = - 30 ` cm The image is 30 cm from the mirror on the same side as the object. ALSO, MAGNIFICATION m = `(-v)/(u) = ((-30))/((-10)) = - `3 The image is magnified. real and inverted . ii.The object distance u = -5 cm . Then from Eq. 6, `(1)/(v) + (1)/(-5) = (1)/(-7.5) or v = (5 x 7.5)/((7.5 - 5))`= 15 cm this image is formed at 15 cm behing the mirror. it is a virtual image. magnification m = `- (v)/(u) = (-15)/((-5)) = 3 ` The image is magnified, virtual and erect. |
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