An object is placed at a distance of 1.50 , m | vedclass

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(A) Given: Total distance $D = u + v = 1.50 \, m$. Magnification $m = -4$ (since the image is real and formed on a screen).Using the magnification for

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$24$

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(A) Given: Total distance $D = u + v = 1.50 \, m$. Magnification $m = -4$ (since the image is real and formed on a screen).Using the magnification formula $m = v/u$,we have $-4 = v/u$,which implies $v = 4u$.Substituting this into the distance equation: $u + 4u = 1.50 \, m$,so $5u = 1.50 \, m$,which gives $u = 0.3 \, m$ and $v = 1.2 \, m$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{1.2} - \frac{1}{-0.3} = \frac{1}{1.2} + \frac{1}{0.3} = \frac{1 + 4}{1.2} = \frac{5}{1.2}$.Therefore,$f = \frac{1.2}{5} = 0.24 \, m = 24 \, cm$.

An object is placed at a distance of $1.50 \, m$ from a screen. $A$ convex lens is placed between them to produce a four times magnified image on the screen. The focal length of the lens is ..... $cm$.

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