The photograph of a house occupies an area of $1.75 c m^{2}$ on a $35 m m$ slide. The slide is projected on to a screen, and the area of the house on the screen is $1.55 m^{2}$ . What is the linear magnification of the projector-screen arrangement.

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Solution

$Arial Magnification = Area of image / Area of object$
$= 1.55 / 1.75 \times 10^{4}$
$= 8857$
$Linear Magnification = \sqrt{8857} = 94.11$

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The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55m2. What is the linear magnification of the projector-screen arrangement.

Arial Magnification = Area of image / Area of object =1.55/1.75×104 =8857Linear Magnification=√8857=94.11

The photograph of a house occupies an area of $1.75 c m^{2}$ on a $35 m m$ slide. The slide is projected on to a screen, and the area of the house on the screen is $1.55 m^{2}$ . What is the linear magnification of the projector-screen arrangement.

$Arial Magnification = Area of image / Area of object$
$= 1.55 / 1.75 \times 10^{4}$
$= 8857$
$Linear Magnification = \sqrt{8857} = 94.11$