An object is placed at a distance of 40 , cm | vedclass

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(B) Using the mirror formula,$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.For a concave mirror,$f = -15 \, cm$.Initially,the object distance $u_1 = -40 \

Vedclass · Accepted Answer

$36 \, cm$ away from the mirror

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(B) Using the mirror formula,$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.For a concave mirror,$f = -15 \, cm$.Initially,the object distance $u_1 = -40 \, cm$.$\frac{1}{-15} = \frac{1}{v_1} + \frac{1}{-40} \Rightarrow \frac{1}{v_1} = \frac{1}{40} - \frac{1}{15} = \frac{3 - 8}{120} = \frac{-5}{120} = -\frac{1}{24}$.So,$v_1 = -24 \, cm$ (image is $24 \, cm$ in front of the mirror).When the object is displaced by $20 \, cm$ towards the mirror,the new object distance $u_2 = -(40 - 20) = -20 \, cm$.$\frac{1}{-15} = \frac{1}{v_2} + \frac{1}{-20} \Rightarrow \frac{1}{v_2} = \frac{1}{20} - \frac{1}{15} = \frac{3 - 4}{60} = -\frac{1}{60}$.So,$v_2 = -60 \, cm$ (image is $60 \, cm$ in front of the mirror).The displacement of the image is $|v_2 - v_1| = |-60 - (-24)| = |-36| = 36 \, cm$.Since the magnitude of the image distance increased,the image shifts $36 \, cm$ away from the mirror.

An object is placed at a distance of $40 \, cm$ from a concave mirror of focal length $15 \, cm$. If the object is displaced through a distance of $20 \, cm$ towards the mirror,the displacement of the image will be:

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