In the displacement method,a convex lens form | vedclass

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(D) In the displacement method,for a fixed distance between the object and the screen,there are two positions of the lens where a real image is formed

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) In the displacement method,for a fixed distance between the object and the screen,there are two positions of the lens where a real image is formed.Let $h$ be the height of the object,$h_1$ be the height of the first image,and $h_2$ be the height of the second image.The magnification for the two positions is given by $m_1 = \frac{h_1}{h}$ and $m_2 = \frac{h_2}{h}$.According to the properties of the displacement method,the product of the magnifications is $m_1 	imes m_2 = 1$.Substituting the expressions for magnification: $\left(\frac{h_1}{h}ight) 	imes \left(\frac{h_2}{h}ight) = 1$.This simplifies to $h^2 = h_1 	imes h_2$,or $h = \sqrt{h_1 h_2}$.Given $h_1 = 24 \,cm$ and $h_2 = 6 \,cm$,we calculate $h = \sqrt{24 	imes 6} = \sqrt{144} = 12 \,cm$.Therefore,the height of the object is $12 \,cm$.

In the displacement method,a convex lens forms a real image of an object for its two different positions. If the heights of the images in the two cases are $24 \,cm$ and $6 \,cm$,then the height of the object is ........ $cm$.

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