An object ' O ' is placed at a distance of 10 | vedclass

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(D) Given: Initial distance of object $u_0 = -100\,cm$. Radius of curvature $R = -200\,cm$. Focal length $f = R/2 = -100\,cm$.Speed of object $v_{obj}

Vedclass · Accepted Answer

$400$

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(D) Given: Initial distance of object $u_0 = -100\,cm$. Radius of curvature $R = -200\,cm$. Focal length $f = R/2 = -100\,cm$.Speed of object $v_{obj} = 2\,cm/s$ towards the mirror.After time $t = 10\,s$,the distance moved by the object is $d = v_{obj} 	imes t = 2\,cm/s 	imes 10\,s = 20\,cm$.The new position of the object from the mirror is $u = u_0 + d = -100\,cm + 20\,cm = -80\,cm$.Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Substituting the values: $\frac{1}{v} + \frac{1}{-80} = \frac{1}{-100}$.$\frac{1}{v} = \frac{1}{80} - \frac{1}{100} = \frac{5 - 4}{400} = \frac{1}{400}$.Therefore,$v = 400\,cm$.

An object ' $O$ ' is placed at a distance of $100\,cm$ in front of a concave mirror of radius of curvature $200\,cm$ as shown in the figure. The object starts moving towards the mirror at a speed of $2\,cm/s$. The position of the image from the mirror after $10\,s$ will be at ...... $cm$.

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