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A small object of height `0.5` cm is placed in front of a convex surface of glass `(mu-1.5)` of radius of curvature 10cm. Find the height of the image formed in glass. |
Answer» According to cartesian sign convention, `u=-30cm, R=+10cm, mu_(1)=1,mu_(2)=1.5` Applying the equation, we get ` (1.5)/(v)=(1)/(-30)=(1.5-1)/(+10)` or `v=90cm` (real image) Let `h_(1)` be the height of the image, then `(h_(i))/(h)=(mu_(1)v)/(mu_(2)u)=((1)(90))/((1.5)(-30))=-2` `rArrh_(i)=-2h_(0)(0.5)=-2(0.5)=-1cm` The negative sign shows that the image is inverted. |
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