An illuminated object and a screen are placed | vedclass

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Question

(A) Let the distance of the object from the lens be $u = -x \, cm$.The distance of the screen from the lens is $v = +(90 - x) \, cm$.The magnification

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) Let the distance of the object from the lens be $u = -x \, cm$.The distance of the screen from the lens is $v = +(90 - x) \, cm$.The magnification $m$ for a real image is given by $m = \frac{v}{u} = -2$.Substituting the values: $\frac{90 - x}{-x} = -2$.$90 - x = 2x \Rightarrow 3x = 90 \Rightarrow x = 30 \, cm$.Thus,$u = -30 \, cm$ and $v = 90 - 30 = 60 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.$\frac{1}{f} = \frac{1}{60} - \frac{1}{-30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20}$.Therefore,$f = 20 \, cm$.

An illuminated object and a screen are placed $90 \, cm$ apart. What is the focal length of the lens required to produce an image on the screen,twice the size of the object?.......$cm$

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