A convex lens is placed 10, cm from a light s | vedclass

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Question

(D) Given,object distance $u = -10\, cm$ and image distance $v = +10\, cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we get $\f

Vedclass · Accepted Answer

$0.55\, cm$ away from the lens

Vedclass · Answer

(D) Given,object distance $u = -10\, cm$ and image distance $v = +10\, cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{10} - \frac{1}{-10} = \frac{1}{f} \Rightarrow \frac{2}{10} = \frac{1}{f} \Rightarrow f = 5\, cm$.When a glass slab of thickness $t = 1.5\, cm$ and refractive index $\mu = 1.5$ is placed in front of the source,the apparent shift in the position of the object is given by $\Delta x = t(1 - \frac{1}{\mu}) = 1.5(1 - \frac{1}{1.5}) = 1.5(1 - \frac{2}{3}) = 1.5(\frac{1}{3}) = 0.5\, cm$.The object effectively moves closer to the lens,so the new object distance is $u' = -(10 - 0.5) = -9.5\, cm$.Using the lens formula again to find the new image position $v'$: $\frac{1}{v'} - \frac{1}{-9.5} = \frac{1}{5} \Rightarrow \frac{1}{v'} = \frac{1}{5} - \frac{1}{9.5} = \frac{9.5 - 5}{47.5} = \frac{4.5}{47.5}$.Thus,$v' = \frac{47.5}{4.5} \approx 10.55\, cm$.The shift in the screen position is $d = v' - v = 10.55 - 10 = 0.55\, cm$ away from the lens.

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