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A linear objectof height 10 cm is kept in front of a concave mirror of radius of curvature 15 cm, at a distance of 10 cm. The image formed is |
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Answer» magnified and erect Radius of curvature (R ) = 15 cm Distance of object (U) `= -10 cm` We KNOW that, `f=(R )/(2)=(15)/(2)cm` According to mirror formula, `(1)/(f)=(1)/(u)+(1)/(v)` `(1)/(-(15)/(2))=(1)/(-10)+(1)/(v)` `rArr (1)/(v)=(-2)/(15)+(1)/(10)` `rArr (1)/(v)=(-4+3)/(30)` `rArr (1)/(v)=(-1)/(30)` `rArr v=-30 cm` MAGNIFICATION (m) `= (h_(i))/(h_(o))=(-v)/(u)` `= (h_(i))/(10)=(-(-30))/(-10)` `h_(i)=-3` |
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