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An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case. |
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Answer» Solution :The focal LENGTH `f=-15//2cm=-7.5cm` (i) The object distance `u=-10cm.` Then Eq. (9.7) GIVES `(1)/(v)+(1)/(-10)=(1)/(-7.5)` `"or"v=(10xx7.5)/(-2.5)=-30CM` 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. (9.7), `(1)/(v)+(1)/(-5)=(1)/(-7.5)` `or v=(5xx7.5)/((7.5-5))=15cm` This image is formed at 15 cm behind 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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